JEE Main 27 July 2022 Shift 2 question paper with solutions

JEE Main 27 July 2022 Shift 2: all 90 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.

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Maths

Question 1

Maths · Inverse Trigonometric Functions · Single correct

The domain of the function \[ f(x)=\sin^{-1}\!\left([2x^2-3]\right)+\log_{2}\!\left(\log_{\frac12}\!\left(x^2-5x+5\right)\right), \] where $[t]$ is the greatest integer function, is:

  1. $(-\\sqrt{\\frac{5}{2}}, \\frac{5-\\sqrt{5}}{2})$
  2. $(\\frac{5-\\sqrt{5}}{2}, \\frac{5+\\sqrt{5}}{2})$
  3. $(1, \\frac{5-\\sqrt{5}}{2})$
  4. $[1, \\frac{5+\\sqrt{5}}{2}]$

Answer: (c)

Solution

Given $f(x) = \sin^{-1}[2x^2 - 3] + \log_2 \left( \log_{\frac{1}{2}} \left( x^2 - 5x + 5 \right) \right)$. $P_1$: $-1 \leq [2x^2 - 3] 0$ $$\Rightarrow \left( x - \frac{5 - \sqrt{5}}{2} \right) \left( x - \frac{5 + \sqrt{5}}{2} \right) > 0$$ $P_3$: $\log_{\frac{1}{2}} \left( x^2 - 5x + 5 \right) > 0$ $$\Rightarrow x^2 - 5x - 5 < 1$$ $$\Rightarrow x^2 - 5x + 4 < 0$$ $$\Rightarrow P_3: x \in (1, 4)$$ So, $P_1 \cap P_2 \cap P_3 = \left( 1, \frac{5 - \sqrt{5}}{2} \right)$

Question 2

Maths · Complex Numbers and Quadratic Equations · Single correct

Let S be the set of all $(\alpha, \beta)$, $\pi < \alpha$, $\beta < 2\pi$, for which the complex number $\frac{1 - i \sin \alpha}{1 + 2 i \sin \alpha}$ is purely imaginary and $\frac{1 + i \cos \beta}{1 - 2 i \cos \beta}$ is purely real. Let $Z_{\alpha \beta} = \sin 2\alpha + i \cos 2\beta$, $(\alpha, \beta) \in S$. Then $$\sum_{(\alpha, \beta) \in S} \left( i Z_{\alpha \beta} + \frac{1}{i\overline{Z_{\alpha\beta}}} \right)$$ is equal to:

  1. 3
  2. 3i
  3. 1
  4. 2 - i

Answer: (c)

Solution

Given $\pi < \alpha, \beta < 2\pi$. $$\frac{1 - i \sin \alpha}{1 + i(2 \sin \alpha)} = Purely imaginary$$ $$\Rightarrow \frac{(1 - i \sin \alpha)(1 - i(2 \sin \alpha))}{1 + 4 \sin^2 \alpha} = Purely imaginary$$ $$\Rightarrow \frac{1 - 2 \sin^2 \alpha}{1 + 4 \sin^2 \alpha} = 0$$ $$\Rightarrow \sin^2 \alpha = \frac{1}{2}$$ $$\Rightarrow \alpha = \left\{ \frac{5\pi}{4}, \frac{7\pi}{4} \right\}$$ And $$\frac{1 + i \cos \beta}{1 + i(-2 \cos \beta)} = Purely real$$ $$\Rightarrow \frac{(1 + i \cos \beta)(1 + 2 \cos \beta)}{1 + 4 \cos^2 \beta} = Purely real$$ $$\Rightarrow 3 \cos \beta = 0$$ $$\Rightarrow \beta = \frac{3\pi}{2}$$ $$\Rightarrow Z_{\alpha \beta} = \sin \frac{5\pi}{2} + i \cos 3\pi = 1 - i$$ or $$Z_{\alpha \beta} = \sin \frac{7\pi}{2} + i \cos 3\pi = -1 - i$$ Required value = $\left$[ $\frac{i(1-i) + \frac{1}{i(1+i)}}{i}$ $\right$] + $\left$[ $\frac{i(-1-i) + \frac{1}{i(-1+i)}}{i}$ $\right$]$$ $$= i(-2i) + $\frac{1}{i}$ $\frac{2i}{(-2)}$ $\Rightarrow$ 2 - 1 = 1

Question 3

Maths · Complex Numbers and Quadratic Equations · Single correct

If $\alpha$, $\beta$ are the roots of the equation $$x^2 - \left(5 + 3 \sqrt[3]{\log_3 5} - 5 \sqrt[3]{\log_5 3}\right) + 3 \left(3^{(\log_3 5)^{\frac{1}{3}}} - 5^{(\log_5 3)^{\frac{2}{3}}} - 1\right) = 0$$ then the equation, whose roots are $$\alpha + \frac{1}{\beta} \text{ and } \beta + \frac{1}{\alpha},$$

  1. $3x^2 - 20x - 12 = 0$
  2. $3x^2 - 10x - 4 = 0$
  3. $3x^2 - 10x + 2 = 0$
  4. $3x^2 - 20x + 16 = 0$

Answer: (b)

Solution

Bonus because 'x' is missing the correct will be, $$x^2 - \left(5 + 3^{\log_3 5} - 5^{\log_5 3}\right)x + 3\left[3^{\left(\log_3 5\right)^{\frac{1}{3}}} - 5^{\left(\log_5 3\right)^{\frac{2}{3}}} - 1\right] = 0$$ $$3^{\frac{1}{\log_3 5}} = 3^{\log_3 5 \cdot \frac{1}{\log_3 5}} = 3^{\log_3 5}$$ $$= (3^{\log_3 5})^{\frac{1}{\log_5 3}} = 5^{\log_5 3}$$ $$3^{\frac{1}{\log_3 5}} = 3^{\log_3 5 \cdot \sqrt[3]{\log_5 3}} = (3^{\log_3 5})^{(\log_5 3)^{2/3}}$$ $$= 5^{(\log_5 3)^{2/3}}$$ So, equation is $x^2 - 5x - 3 = 0$ and roots are $\alpha$ and $\beta$ $\{\alpha + \beta = 5; \alpha \beta = -3\}$ New roots are $\alpha + \frac{1}{\beta}$ and $\beta + \frac{1}{\alpha}$ i.e., $\frac{\alpha \beta + 1}{\beta}$ and $\frac{\alpha \beta + 1}{\alpha}$ i.e., $\frac{-2}{\beta}$ and $\frac{-2}{\alpha}$ Let $\frac{-2}{\alpha} = t \implies \alpha = \frac{-2}{t}$ As $\alpha^2 - 5\alpha - 3 = 0$ $$\implies \left(\frac{-2}{t}\right)^2 - 5\left(\frac{-2}{t}\right) - 3 = 0$$ $$\implies \frac{4}{t^2} + \frac{10}{t} - 3 = 0$$ $$\implies 4 + 10t - 3t^2 = 0$$ $$\implies 3t^2 - 10t - 4 = 0$$ i.e., $3x^2 - 10x - 4 = 0$

Question 4

Maths · Matrices · Single correct

Let $A = \begin{pmatrix} 4 & -2 \\ \alpha & \beta \end{pmatrix}$ If $A^2 + \gamma A + 18I = O$, then $\det(A)$ is equal to

  1. -18
  2. 18
  3. -50
  4. 50

Answer: (b)

Solution

The characteristic equation for $A$ is $|A - \lambda I| = 0$. $$\begin{vmatrix} 4 - \lambda & -2 \\ \alpha & \beta - \lambda \end{vmatrix} = 0$$ This implies $$(4 - \lambda)(\beta - \lambda) + 2\alpha = 0$$ Expanding, we get $$\lambda^2 - (\beta + 4)\lambda + 4\beta + 2\alpha = 0$$ Put $\lambda = A$. Then, $$A^2 - (\beta + 4)A + (4\beta + 2\alpha)I = 0$$ On comparison, $$-9(\beta + 4) = \gamma \& 4\beta + 2\alpha = 18$$ and $$|A| = 4\beta + 2\alpha = 18$$

Question 5

Maths · Continuity and Differentiability · Single correct

If for $p \neq q \neq 0$, then function $$f(x) = \frac{\sqrt[7]{p(729 + x)} - 3}{\sqrt[3]{729 + qx} - 9}$$ is continuous at $x = 0$, then:

  1. $7pq \, f(0) - 1 = 0$
  2. $63q \, f(0) - p^2 = 0$
  3. $21q \, f(0) - p^2 = 0$
  4. $7pq \, f(0) - 9 = 0$

Answer: (b)

Solution

Given $f(0) = \lim_{x \to 0} f(x)$. Limit should be $\frac{0}{0}$ form. So, $\sqrt[7]{p \cdot 729} - 3 = 0 \implies p \cdot 3^6 = 3^7 \implies p = 3$. Now, $f(0) = \lim_{x \to 0} \frac{\sqrt[7]{3(3^6 + x)} - 3}{\sqrt[3]{3^6} + qx - 9}$. $$= \lim_{x \to 0} \frac{3 \left[ \left( 1 + \frac{x}{3^6} \right)^{1/7} - 1 \right]}{9 \left[ \left( 1 + \frac{qx}{3^6} \right)^{1/3} - 1 \right]} = \frac{3}{9} \times \frac{1}{q} \frac{7 \cdot 3^6}{3 \cdot 3^6}$$ $$\implies f(0) = \frac{1}{3} \times \frac{3}{7q} = \frac{1}{7q}$$ $$\implies 7qf(0) - 1 = 0$$ $$\implies 7 \cdot p^2 \cdot qf(0) - p^2 = 0 (for option)$$ $$\implies 63qf(0) - p^2 = 0$$

Question 6

Maths · Applications of Integrals · Single correct

Let $f(x) = 2 + |x| - |x - 1| + |x + 1|$, $x \in \mathbb{R}$. Consider (S1): $f'\left(-\frac{3}{2}\right) + f'\left(-\frac{1}{2}\right) + f'\left(\frac{1}{2}\right) + f'\left(\frac{3}{2}\right) = 2$ (S2): $\int_{-2}^{2} f(x) \, dx = 12$ Then,

  1. both (S1) and (S2) are correct
  2. both (S1) and (S2) are wrong
  3. only (S1) is correct
  4. only (S2) is correct

Answer: (d)

Solution

(S1): f'$\left$(-$\frac{3}{2}$$\right$) + f'$\left$(-$\frac{1}{2}$$\right$) + f'$\left$($\frac{1}{2}$$\right$) + f'$\left$($\frac{3}{2}$$\right$) = 4 (S2): $\int$_{-2}^{2} f(x) $\,$ dx = 12 $\therefore$ (D)

Question 7

Maths · Sequences and Series · Single correct

Let the sum of an infinite G.P., whose first term is $a$ and the common ratio is $r$, be $5$. Let the sum of its first five terms be $\frac{98}{25}$. Then the sum of the first $21$ terms of an AP, whose first term is $10ar$, $n^{th}$ term is $a_n$ and the common difference is $10ar^2$, is equal to:

  1. 21 a_{11}
  2. 22 a_{11}
  3. 15 a_{16}
  4. 14 a_{16}

Answer: (a)

Solution

Given $$S_{21} = \frac{21}{2} [20ar + 20 \cdot 10 ar^2]$$ Simplifying, we have $$= 21 [10 ar + 100 ar^2]$$ Therefore, $$= 21 \cdot a_{11}$$

Question 8

Maths · Applications of Integrals · Single correct

The area of the region enclosed by $y \leq 4x^2$, $x^2 \leq 9y$ and $y \leq 4$, is equal to:

  1. $\frac{40}{3}$
  2. $\frac{56}{3}$
  3. $\frac{112}{3}$
  4. $\frac{80}{3}$

Answer: (d)

Solution

The area $\Delta$ is calculated as follows: $$\Delta = 2 \cdot \int_0^4 \left( 3 \sqrt{y} - \frac{\sqrt{y}}{2} \right) \, dy$$ Simplifying the integrand: $$= 2 \cdot \int_0^4 \frac{5}{2} \sqrt{y} \, dy = \frac{80}{3}$$

Question 9

Maths · Integrals · Single correct

\[ \int_{0}^{2} \left( \left|2x^2-3x\right| +\left[x-\frac{1}{2}\right] \right)\,dx, \] where \([t]\) is the greatest integer function, is equal to:

  1. $\frac{7}{6}$
  2. $\frac{19}{12}$
  3. $\frac{31}{12}$
  4. $\frac{3}{2}$

Answer: (b)

Solution

Evaluate the integral from 0 to 2 of the absolute value of $2x^2 - 3x$ with respect to $x$. $$\int_0^2 |2x^2 - 3x| \, dx$$ This can be split into two integrals: $$= \int_0^{\frac{3}{2}} (3x - 2x^2) \, dx + \int_{\frac{3}{2}}^2 (2x^2 - 3x) \, dx = \frac{19}{12}.$$ Now consider the integral from 0 to 2 of $\left[x - \frac{1}{2}\right]$ with respect to $x$: $$\int_0^2 \left[x - \frac{1}{2}\right] \, dx = \int_{-\frac{1}{2}}^{\frac{3}{2}} [t] \, dt$$ This can be evaluated as: $$= \int_{-\frac{1}{2}}^0 (-1) \, dt + \int_0^1 0 \cdot dt + \int_1^{\frac{3}{2}} 1 \cdot dt = 0.$$

Question 10

Maths · Applications of Derivatives · Single correct

Consider a curve $y = y(x)$ in the first quadrant as shown in the figure. Let the area $A_1$ is twice the area $A_2$. Then the normal to the curve perpendicular to the line $2x - 12y = 15$ does NOT pass through the point.

  1. (6, 21)
  2. (8, 9)
  3. (10, -4)
  4. (12, -15)

Answer: (c)

Solution

Given that $A_1 = 2A_2$ from the graph $A_1 + A_2 = xy - 8$ $$\Rightarrow \frac{3}{2} A_1 = xy - 8$$ $$\Rightarrow A_1 = \frac{2}{3} xy - \frac{16}{3}$$ $$\Rightarrow \int_4^x f(x) \, dx = \frac{2}{3} xy - \frac{16}{3}$$ $$\Rightarrow f(x) = \frac{2}{3} \left( x \frac{dy}{dx} + y \right)$$ $$\Rightarrow \frac{2}{3} x \frac{dy}{dx} = \frac{y}{3}$$ $$\Rightarrow 2 \int \frac{dy}{y} = \int \frac{dx}{x}$$ $$\Rightarrow 2 \ln y = \ln x + \ln c$$ $$\Rightarrow y^2 = cx$$ As $f(4) = 2 \Rightarrow c = 1$ so $y^2 = x$ slope of normal $= -6$ $$y = -6(x) - \frac{1}{2}(-6) - \frac{1}{4}(-6)^3$$ $$\Rightarrow y = -6x + 3 + 54$$ $$\Rightarrow y + 6x = 57$$ Now check options and (C) will not satisfy.

Question 11

Maths · Determinants · Single correct

The equations of the sides AB, BC and CA of a triangle ABC are $2x + y = 0$, $x + py = 39$ and $x - y = 3$ respectively and $P(2, 3)$ is its circumcentre. Then which of the following is NOT true :

  1. $(AC)^2 = 9p$
  2. $(AC)^2 + p^2 = 136$
  3. $32 < area (\Delta ABC) < 36$
  4. $34 < area (\Delta ABC) < 38$

Answer: (d)

Solution

Perpendicular bisector of AB is given by $x + y = 5$. Take the image of A. $$\frac{x - 1}{1} = \frac{y + 2}{1} = \frac{-2(-6)}{2} = 6$$ The coordinates are $(7, 4)$. Solving $7 + 4p = 39$, we find $p = 8$. Solving $x + 8y = 39$ and $y = -2x$, we get $$x = \frac{-39}{15}, y = \frac{78}{15}$$ For $AC^2 = 72 = 9p$, we have $$AC^2 + p^2 = 72 + 64 = 136$$ The area of $\Delta ABC$ is given by $$\Delta ABC = \frac{1}{2} \begin{vmatrix} 1 & -2 & 1 \\ 7 & 4 & 1 \\ -\frac{39}{15} & \frac{78}{15} & 1 \end{vmatrix}$$ $$= \frac{1}{2} \left[ 18 + 18 \times \frac{13}{5} \right]$$ $$= 9 \left[ \frac{18}{5} \right] = \frac{162}{5} = 32.4$$

Question 12

Maths · Conic Sections · Single correct

A circle $C_1$ passes through the origin $O$ and has diameter 4 on the positive $x$-axis. The line $y = 2x$ gives a chord $OA$ of a circle $C_1$. Let $C_2$ be the circle with $OA$ as a diameter. If the tangent to $C_2$ at the point $A$ meets the $x$-axis at $P$ and $y$-axis at $Q$, then $QA : AP$ is equal to:

  1. 1 : 4
  2. 1 : 5
  3. 2 : 5
  4. 1 : 3

Answer: (a)

Solution

Given $C_1: x^2 + y^2 - 4x = 0$ and $\tan \theta = 2$. The line $L: 2x - y = 0$ is tangent to the circle $C_1$ at point $A$. The circle $C_2$ is a circle with $OA$ as diameter. So, the tangent at $A$ on $C_2$ is perpendicular to $OR$. Let $OA = \ell$. Therefore, $$\frac{QA}{AP} = \frac{\ell \cot \theta}{\ell \tan \theta} = \frac{1}{\tan^2 \theta} = \frac{1}{4}.$$

Question 13

Maths · Conic Sections · Single correct

If the length of the latus rectum of a parabola, whose focus is $(a, a)$ and the tangent at its vertex is $x + y = a$, is 16, then $|a|$ is equal to:

  1. $2\sqrt{2}$
  2. $2\sqrt{3}$
  3. $4\sqrt{2}$
  4. 4

Answer: (c)

Solution

Given the equation $x + y = a$. The absolute value of $|P|$ is given by $$|P| = \left| \frac{a}{\sqrt{2}} \right| = \frac{16}{4} = 4.$$ Therefore, $$|a| = 4\sqrt{2}.$$

Question 14

Maths · Three Dimensional Geometry · Single correct

If the length of the perpendicular drawn from the point $P(a, 4, 2)$, $a > 0$ on the line $$\frac{x+1}{2} = \frac{y-3}{3} = \frac{z-1}{-1}$$ is $2\sqrt{6}$ units and $Q(\alpha_1, \alpha_2, \alpha_3)$ is the image of the point $P$ in this line, then $a + \sum_{i=1}^{3} \alpha_i$ is equal to:

  1. 7
  2. 8
  3. 12
  4. 14

Answer: (b)

Solution

(2$\lambda$ - 1 - a)2 + (3$\lambda$ - 1)3 + (-$\lambda$ - 1)(-1) = 0 $\Rightarrow$ 4$\lambda$ - 2 - 2a + 9$\lambda$ - 3 + $\lambda$ + 1 = 0 $\Rightarrow$ 14$\lambda$ - 4 - 2a = 0 $\Rightarrow$ 7$\lambda$ - 2 - a = 0 and, (2$\lambda$ - 1 - a)^2 + (3$\lambda$ - 1)^2 + ($\lambda$ + 1)^2 = 24 $\Rightarrow$ (5$\lambda$ - 1)^2 + (3$\lambda$ - 1)^2 + ($\lambda$ + 1)^2 = 24 $\Rightarrow$ 35$\lambda$^2 - 14$\lambda$ - 21 = 0 $\Rightarrow$ ($\lambda$ - 1)(35$\lambda$ + 21) = 0 For, $\lambda$ = 1 $\Rightarrow$ a = 5 Let ($\alpha$_1, $\alpha$_2, $\alpha$_3) be reflection of point P $\alpha$_1 + 5 = 2 $\alpha$_2 + 4 = 12 $\alpha$_3 + 2 = 0 $\alpha$_1 = -3 $\alpha$_2 = 8 $\alpha$_3 = -2 a + $\alpha$_1 + $\alpha$_2 + $\alpha$_3 = 8

Question 15

Maths · Three Dimensional Geometry · Single correct

If the line of intersection of the planes $ax + by = 3$ and $ax + by + cz = 0$, $a > 0$ makes an angle $30^\circ$ with the plane $y - z + 2 = 0$, then the direction cosines of the line are:

  1. $\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0$
  2. $\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0$
  3. $\frac{1}{\sqrt{5}}, -\frac{2}{\sqrt{5}}, 0$
  4. $\frac{1}{2}, -\frac{\sqrt{3}}{2}, 0$

Answer: (b)

Solution

Given $\mathbf{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a & b & 0 \\ a & b & c \end{vmatrix}$ $= bc\hat{i} - ac\hat{j}$ Direction ratios of line are $(b,\ -a,\ 0)$ Direction ratios of normal of the plane are $(0,\ 1,\ -1)$ $$\cos 60^\circ = \left|\frac{-a}{\sqrt{2}\sqrt{b^2 + a^2}}\right| = \frac{1}{2}$$ $$\Rightarrow \left|\frac{a}{\sqrt{a^2 + b^2}}\right| = \frac{1}{\sqrt{2}}$$ $\Rightarrow b = \pm\, a$ So, D.R.'s can be $(\pm\, a,\ -a,\ 0)$ $\therefore$ D.C.'s can be $\pm\left(\dfrac{\pm 1}{\sqrt{2}},\ -\dfrac{1}{\sqrt{2}},\ 0\right)$

Question 16

Maths · Probability · Single correct

Let X have a binomial distribution B(n, p) such that the sum and the product of the mean and variance of X are 24 and 128 respectively. If P(X > n - 3) = $\frac{k}{2^n}$, then k is equal to

  1. 528
  2. 529
  3. 629
  4. 630

Answer: (b)

Solution

Let $\alpha = Mean$ and $\beta = Variance$ ($\alpha > \beta$). So, $\alpha + \beta = 24$, $\alpha \beta = 128$. Therefore, $\alpha = 16$ and $\beta = 8$. Thus, $np = 16$ and $npq = 8 \implies q = \frac{1}{2}$. Therefore, $p = \frac{1}{2}$, $n = 32$. The probability $p(x > n - 3) = \frac{1}{2^n} \left( ^nC_{n-2} + ^nC_{n-1} + ^nC_n \right)$. Therefore, $k = ^{32}C_{30} + ^{32}C_{31} + ^{32}C_{32} = \frac{32 \times 31}{2} + 32 + 1 = 496 + 33 = 529$.

Question 17

Maths · Probability · Single correct

A six faced die is biased such that $3 \times \mathrm{P}$(a prime number) $= 6 \times \mathrm{P}$(a composite number) = $2 \times \mathrm{P}(1)$. Let $X$ be a random variable that counts the number of times one gets a perfect square on some throws of this die. If the die is thrown twice, then the mean of $X$ is:

  1. $\frac{3}{11}$
  2. $\frac{5}{11}$
  3. $\frac{7}{11}$
  4. $\frac{8}{11}$

Answer: (d)

Solution

Let $\dfrac{P(\text{a prime number})}{2}=k$ $\dfrac{P(\text{a composite number})}{1}=k$ $\dfrac{P(1)}{3}=k$. So, $\($ P(a prime number) = 2k $\)$, $\($ P(a composite number) = k $\)$, and $\($ P(1) = 3k $\)$. And $\($ 3 $\times$ 2k + 2 $\times$ k + 3k = 1 $\)$. $\($ $\Rightarrow$ k = $\frac{1}{11}$ $\)$. $\($ P(success) = P(1 or 4) = 3k + k = $\frac{4}{11}$ $\)$. Number of trials, $\($ n = 2 $\)$. Therefore, mean $\($ = np = 2 $\times$ $\frac{4}{11}$ = $\frac{8}{11}$ $\)$.

Question 18

Maths · Heights and Distances · Single correct

The angle of elevation of the top P of a vertical tower PQ of height 10 from a point A on the horizontal ground is $45^\circ$. Let R be a point on AQ and from a point B, vertically above R, the angle of elevation of P is $60^\circ$. If $\angle BAQ = 30^\circ$, AB = d and the area of the trapezium PQRB is $\alpha$, then the ordered pair (d, $\alpha$) is:

  1. $(10(\\sqrt{3} - 1), 25)$
  2. $(10(\\sqrt{3} - 1), \\frac{25}{2})$
  3. $(10(\\sqrt{3} + 1), 25)$
  4. $(10(\\sqrt{3} + 1), \\frac{25}{2})$

Answer: (a)

Solution

QA = 10 RA = d $\cos$ 30^$\circ$ = $\frac{\sqrt{3}d}{2}$ $\tan$ 60^$\circ$ = $\frac{PQ - BR}{QR}$ = $\frac{10 - \frac{d}{2}}{10 - \frac{\sqrt{3}d}{2}}$ $\Rightarrow$ $\sqrt{3}$ = $\frac{20 - d}{20 - \sqrt{3}d}$ $\Rightarrow$ 20$\sqrt{3}$ - 3d = 20 - d $\Rightarrow$ 2d = 20($\sqrt{3}$ - 1) $\Rightarrow$ d = 10($\sqrt{3}$ - 1) ar(PQRB) = $\alpha$ = $\frac{1}{2}$ (PQ + BR) $\cdot$ QR = $\frac{1}{2}$ $\left$(10 + $\frac{d}{2}$$\right$) $\cdot$ $\left$(10 - $\frac{\sqrt{3}d}{2}$$\right$) = $\frac{1}{2}$ (10 + 5$\sqrt{3}$ - 5)(10 - 15 + 5$\sqrt{3}$) = $\frac{1}{2}$ (5$\sqrt{3}$ + 5)(5$\sqrt{3}$ - 5) = $\frac{1}{2}$ (75 - 25) = 25

Question 19

Maths · Trigonometric Functions · Single correct

Let $S = \left\{ \theta \in \left(0, \frac{\pi}{2}\right) : \sum_{m=1}^{9} \sec \left(\theta + (m-1) \frac{\pi}{6}\right) \sec \left(\theta + \frac{m\pi}{6}\right) = -\frac{8}{\sqrt{3}} \right\}$ Then

  1. $S = \left\{ \frac{\pi}{12} \right\}$
  2. $S = \left\{ \frac{2\pi}{3} \right\}$
  3. $\sum_{\theta \in S} \theta = \frac{\pi}{2}$
  4. $\sum_{\theta \in S} \theta = \frac{3\pi}{4}$

Answer: (c)

Solution

Let $\alpha = \theta + (m-1) \frac{\pi}{6}$ and $\beta = \theta + m \frac{\pi}{6}$. So, $\beta - \alpha = \frac{\pi}{6}$. Here, $$\sum_{m=1}^{9} \sec \alpha \cdot \sec \beta = \sum_{m=1}^{9} \frac{1}{\cos \alpha \cdot \cos \beta}$$ $$= 2 \sum_{m=1}^{9} \frac{\sin(\beta - \alpha)}{\cos \alpha \cdot \cos \beta} = 2 \sum_{m=1}^{9} (\tan \beta - \tan \alpha)$$ $$= 2 \sum_{m=1}^{9} \left( \tan \left( \theta + m \frac{\pi}{6} \right) - \tan \left( \theta + (m-1) \frac{\pi}{6} \right) \right)$$ $$= 2 \left( \tan \left( \theta + \frac{9\pi}{6} \right) - \tan \theta \right) = 2 (-\cot \theta - \tan \theta) = -\frac{8}{\sqrt{3}}$$ (Given) Therefore, $\tan \theta + \cot \theta = \frac{4}{\sqrt{3}}$. $$\Rightarrow \tan \theta = \frac{1}{\sqrt{3}} or \sqrt{3}$$ So, $S = \left\{ \frac{\pi}{6}, \frac{\pi}{3} \right\}$. $$\sum_{\theta \in S} \theta = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$$

Question 20

Maths · Mathematical Reasoning · Single correct

If the truth value of the statement $\left( P \land (\sim R) \right) \rightarrow \left( (\sim R) \land Q \right)$ is F, then the truth value of which of the following is F?

  1. $P \lor Q \rightarrow R$
  2. $R \lor Q \rightarrow \sim P$
  3. $\sim (P \lor Q) \rightarrow R$
  4. $\sim (R \lor Q) \rightarrow \sim P$

Answer: (d)

Solution

Given $X \Rightarrow Y$ is false when $X$ is true and $Y$ is false. So, $P \rightarrow T$, $Q \rightarrow F$, $R \rightarrow F$. (A) $P \lor Q \rightarrow \sim R$ is T (B) $R \lor Q \rightarrow \sim P$ is T (C) $\sim (P \lor Q) \rightarrow \sim R$ is T (D) $\sim (R \lor Q) \rightarrow \sim P$ is F

Question 21

Maths · Matrices · Numerical

Consider a matrix $A= \begin{bmatrix} \alpha & \beta & \gamma\\ \alpha^2 & \beta^2 & \gamma^2\\ \beta+\gamma & \gamma+\alpha & \alpha+\beta \end{bmatrix}$ where $\alpha,\beta,\gamma$ are three distinct natural numbers. If $\frac{\det(\operatorname{adj}(\operatorname{adj}(\operatorname{adj}(A))))} {(\alpha-\beta)^{16}(\beta-\gamma)^{16}(\gamma-\alpha)^{16}} =2^{32}\times3^{16}$, then the number of such $3$-tuples $(\alpha,\beta,\gamma)$ is ______.

Answer: 42

Solution

Given the matrix $A = \begin{bmatrix} \alpha & \beta & \gamma \\ \alpha^2 & \beta^2 & \gamma^2 \\ \beta + \gamma & \gamma + \alpha & \alpha + \beta \end{bmatrix}$. Perform the row operation $R_3 \rightarrow R_3 + R_1$. This gives: $$|A| = |\alpha + \beta + \gamma| \begin{vmatrix} \alpha & \beta & \gamma \\ \alpha^2 & \beta^2 & \gamma^2 \\ 1 & 1 & 1 \end{vmatrix}$$ Therefore, $$|A| = (\alpha + \beta + \gamma)(\alpha - \beta)(\beta - \gamma)(\gamma - \alpha)$$ The adjugate of $A$ is given by $|\mathrm{adj} \, A| = |A|^{n-1}$. Thus, $$|\mathrm{adj} \, (\mathrm{adj} \, A)| = |A|^{(n-1)^2}$$ $$|\mathrm{adj} \, (\mathrm{adj} \, (\mathrm{adj} \, A)))| = |A|^{(n-1)^4} = |A|^{24} = |A|^{16}$$ Therefore, $$(\alpha + \beta + \gamma)^{16} = 2^{32} \cdot 3^{16}$$ $$\Rightarrow (\alpha + \beta + \gamma)^{16} = (2^2 \cdot 3)^{16} = (12)^{16}$$ $$\Rightarrow \alpha + \beta + \gamma = 12$$ Given $\alpha, \beta, \gamma \in \mathbb{N}$, we have $$ (\alpha - 1)+(\beta - 1)+(\gamma - 1)=9 $$ Number of all tuples $(\alpha,\beta,\gamma)$ $=\binom{11}{2}$ $=55$ There is $1$ case for $\alpha=\beta=\gamma$ and $12$ cases when any two of $\alpha,\beta,\gamma$ are equal. Therefore, $\text{No. of distinct tuples }(\alpha,\beta,\gamma)$ $=55-13$ $=42$

Question 22

Maths · Relations and Functions · Fill in the blank

The number of functions $f$, from the set $$A = \{ x \in \mathbb{N} : x^2 - 10x + 9 \leq 0 \}$$ to the set $$B = \{ n^2 : n \in \mathbb{N} \}$$ such that $f(x) \leq (x-3)^2 + 1$, for every $x \in A$, is .

Answer: 1440

Solution

$(x^2 - 10x + 9) \leq 0 \Rightarrow (x-1)(x-9) \leq 0 \Rightarrow x \in [1,9]$ $\Rightarrow A = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ $f(x) \leq (x-3)^2 + 1$ $x = 1 : f(1) \leq 5 \Rightarrow 1^2,\ 2^2$ $x = 2 : f(2) \leq 2 \Rightarrow 1^2$ $x = 3 : f(3) \leq 1 \Rightarrow 1^2$ $x = 4 : f(4) \leq 2 \Rightarrow 1^2$ $x = 5 : f(5) \leq 5 \Rightarrow 1^2,\ 2^2$ $x = 6 : f(6) \leq 10 \Rightarrow 1^2,\ 2^2,\ 3^2$ $x = 7 : f(7) \leq 17 \Rightarrow 1^2,\ 2^2,\ 3^2,\ 4^2$ $x = 8 : f(8) \leq 26 \Rightarrow 1^2,\ 2^2,\ 3^2,\ 4^2,\ 5^2$ $x = 9 : f(9) \leq 37 \Rightarrow 1^2,\ 2^2,\ 3^2,\ 4^2,\ 5^2,\ 6^2$ Total number of such functions $= 2(6!) = 2(720) = 1440$

Question 23

Maths · Binomial Theorem · Numerical

Let for the $9^{\text{th}}$ term in the binomial expansion of $(3 + 6x)^n$, in the increasing powers of $6x$, to be the greatest for $x = \frac{3}{2}$, the least value of n is $n_0$. If k is the ratio of the coefficient of $x^6$ to the coefficient of $x^3$, then $k + n_0$ is equal to:

Answer: 24

Solution

Given $(3 + 6x)^n = \binom{n}{0} 3^n + \binom{n}{1} 3^{n-1} (6x)^1 + \ldots$ $T_{r+1} = \binom{n}{r} 3^{n-r} (6x)^r = \binom{n}{r} 3^{n-r} 6^r \cdot x^r$ $$= \binom{n}{r} 3^n \cdot 3^{-r} \cdot 2^r \cdot \left(\frac{3}{2}\right)^r = \binom{n}{r} 3^n \cdot 3^r$$ [for $x = \frac{3}{2}$] $T_9$ is greatest if $x = \frac{3}{2}$ So, $T_9 > T_{10}$ and $T_9 > T_8$ (concept of numerically greatest term) Here, $\frac{T_9}{T_{10}} > 1$ and $\frac{T_9}{T_8} > 1$ $$\Rightarrow \frac{\binom{n}{8} 3^n 3^8}{\binom{n}{9} 3^n 3^9} > 1 and \frac{\binom{n}{8} 3^n 3^8}{\binom{n}{7} 3^n 3^7} > 1$$ and $\frac{\binom{n}{8}}{\binom{n}{7}} > \frac{1}{3}$ $$\Rightarrow \frac{n-7}{8} > \frac{1}{3}$$ $$\Rightarrow \frac{29}{3} < n < 11 \Rightarrow n = 10 = n_0$$ So, in $(3 + 6x)^n$ for $n = n_0 = 10$ i.e., in $(3 + 6x)^{10}$, here $T_{r+1} = \binom{10}{r} 3^{10-r} 6^r x^r$ $T_7 = \binom{10}{6} 3^4 \cdot 6^6 x^6 = 210 \cdot 3^{10} \cdot 2^6 x^6$ $T_4 = \binom{10}{3} 3^7 \cdot 6^3 x^3 = 120 \cdot 3^{10} \cdot 2^3 x^3$ Ratio of coefficient of $x^6$ and coefficient of $x^3 = k$ $$\therefore k = \frac{210 \cdot 3^{10} \cdot 2^6}{120 \cdot 3^{10} \cdot 2^3} = \frac{7}{4} \times 2^3 = 14$$ So, $k + n_0 = 14 + 10 = 24$

Question 24

Maths · Sequences and Series · Fill in the blank

$$\frac{2^3 - 1^3}{1 \times 7} + \frac{4^3 - 3^3 + 2^3 - 1^3}{2 \times 11} + \frac{6^3 - 5^3 + 4^3 - 3^3 + 2^3 - 1^3}{3 \times 15} + \ldots + \frac{30^3 - 29^3 + 28^3 - 27^3 + \ldots + 2^3 - 1^3}{15 \times 63}$$ is equal to ______.

Answer: 120

Solution

Given $$T_n = \frac{2 \sum_{r=1}^{n} (2r)^3 - \left( \sum_{r=1}^{2n} r^3 \right)}{n(4n+3)}$$ This implies $$T_n = n$$ So, $$\sum_{n=1}^{15} T_n = 120$$

Question 25

Maths · Applications of Derivatives · Numerical

A water tank has the shape of a right circular cone with axis vertical and vertex downwards. Its semi-vertical angle is $\tan^{-1} \frac{3}{4}$. Water is poured in it at a constant rate of 6 cubic meter per hour. The rate (in square meter per hour), at which the wet curved surface area of the tank is increasing, when the depth of water in the tank is 4 meters, is ________.

Answer: 5

Solution

Given $\tan \theta = \frac{3}{4} = \frac{r}{h}$ and $\frac{dV}{dt} = 6$. The volume $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi h^3 \tan^2 \theta = \frac{9 \pi}{48} h^3 = \frac{3 \pi}{16} h^3$. Therefore, $$\frac{dV}{dt} = \frac{3 \pi}{16} \cdot 3h^2 \cdot \frac{dh}{dt} = 6 \implies \left( \frac{dh}{dt} \right)_{h=4} = \frac{2}{3 \pi} m/hr$$ Now, $S = \pi r \ell = \frac{15}{16} \pi h^2$. Thus, $$\frac{dS}{dt} = \frac{15 \pi}{16} \cdot 2h \cdot \frac{dh}{dt}$$ Therefore, $$\left( \frac{dS}{dt} \right)_{h=4} = 5 m^2/hr$$

Question 26

Maths · Continuity and Differentiability · Numerical

For the curve C : $(x^2 + y^2 - 3) + (x^2 - y^2 - 1)^5 = 0$, the value of $3y' - y^3y''$, at the point $(\alpha, \alpha)$, $\alpha > 0$, on C, is equal to .

Answer: 16

Solution

($\alpha$, $\alpha$) lies on C : x^2 + y^2 - 3 + x^2 - y^2 - 1 = 0 Put ($\alpha$, $\alpha$), 2$\alpha$^2 - 3 + 1 = 0 $\Rightarrow$ $\alpha$ = $\sqrt{2}$ Now, differentiate C 2x + 2y $\cdot$ y' + 5(x^2 - y^2 - 1)^4 (2x - 2yy') = 0 $\ldots$ (1) At $\left$( $\sqrt{2}$, $\sqrt{2}$ $\right$) $\sqrt{2}$ + $\sqrt{2}$y' + 5(-1)^4 $\left$( $\sqrt{2}$ - $\sqrt{2}$y' $\right$) = 0 $\Rightarrow$ y' = $\frac{3}{2}$ $\ldots$ (2) Diff. (1) w.r.t. x Again, Diff. (1) w.r.t. x 1 + (y')^2 + yy'' + 20(x^2 - y^2 - 1)^3 (x - yy')^2 $\cdot$ 2 + 5(x^2 - y^2 - 1)^4 $\left$( 1 - (y')^2 - yy'' $\right$) = 0 At $\left$( $\sqrt{2}$, $\sqrt{2}$ $\right$) and y' = $\frac{3}{2}$ We have, $\left$( 1 + $\frac{9}{4}$ $\right$) + $\sqrt{2}$y'' - 40 $\left$( $\sqrt{2}$ - $\sqrt{2}$ $\cdot$ $\frac{3}{2}$ $\right$)^2 + 5(1) $\left$( 1 - $\frac{9}{4}$ - $\sqrt{2}$y'' $\right$) = 0 $\Rightarrow$ 4$\sqrt{2}$y'' = -23 $\therefore$ 3y' - y^3 y'' = $\frac{9}{2}$ + $\frac{23}{2}$ = 16

Question 27

Maths · Integrals · Fill in the blank

Let $f(x)=\min\{[x-1],[x-2],\ldots,[x-10]\}$, where $[t]$ denotes the greatest integer $\le t$. Then $$ \int_{0}^{10} f(x)\,dx + \int_{0}^{10} (f(x))^{2}\,dx + \int_{0}^{10} |f(x)|\,dx $$ is equal to ______.

Answer: 385

Solution

Given $f(x) = [x] - 10$. Evaluate the integral from 0 to 10 of $f(x) \, dx$. $$\int_0^{10} f(x) \, dx = -10 - 9 - 8 - \ldots - 1$$ $$= -\frac{10 \cdot 11}{2} = -55$$ Evaluate the integral from 0 to 10 of $(f(x))^2 \, dx$. $$\int_0^{10} (f(x))^2 \, dx = 10^2 + 9^2 + 8^2 + \ldots + 1^2$$ $$= \frac{10 \cdot 11 \cdot 21}{6} = 385$$ Evaluate the integral from 0 to 10 of $|f(x)| \, dx$. $$\int_0^{10} |f(x)| \, dx = 10 + 9 + 8 + \ldots + 1$$ $$= \frac{10 \cdot 11}{2} = 55$$ Combine the results: $$-55 + 385 + 55 = 385$$

Question 28

Maths · Integrals · Numerical

Let f be a differentiable function satisfying $$f(x) = \frac{2}{\sqrt{3}} \int_{0}^{\sqrt{3}} f\left(\frac{\lambda^2 x}{3}\right) d\lambda, x > 0 and f(1) = \sqrt{3}.$$ If $y = f(x)$ passes through the point $(\alpha, 6)$, then $\alpha$ is equal to _____.

Answer: 12

Solution

Let, $\frac{\lambda^2 x}{3} = t$ $$\Rightarrow \frac{2 \lambda x}{3} \, d\lambda = dt$$ $$\Rightarrow d\lambda = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{x}} \cdot \frac{dt}{\sqrt{t}}$$ So, $f(x) = \frac{1}{\sqrt{x}} \int_0^x \frac{f(t)}{\sqrt{t}} \, dt$ $$\Rightarrow \sqrt{x} \cdot f'(x) + \frac{f(x)}{2\sqrt{x}} = \frac{f(x)}{\sqrt{x}}$$ $$\Rightarrow \sqrt{x} \cdot f'(x) = \frac{f(x)}{2\sqrt{x}}$$ $$\Rightarrow \frac{dy}{y} = \frac{dx}{2x}$$ $$\Rightarrow \ln y = \frac{1}{2} \ln x + c \Rightarrow f(x) = \sqrt{x}$$ $$\Rightarrow y = \sqrt{3x} \{ as f(1) = \sqrt{3} \}$$ So, $f(x) = \sqrt{3x}$ Now, $f(\alpha) = 6 \Rightarrow 36 = 3\alpha$ $$\Rightarrow \alpha = 12$$

Question 29

Maths · Conic Sections · Numerical

A common tangent T to the curves $C_1: \frac{x^2}{4} + \frac{y^2}{9} = 1$ and $C_2: \frac{x^2}{42} - \frac{y^2}{143} = 1$ does not pass through the fourth quadrant. If T touches $C_1$ at $(x_1, y_1)$ and $C_2$ at $(x_2, y_2)$, then $|2x_1 + x_2|$ is equal to ______.

Answer: 20

Solution

Let common tangents are $T_1: y = mx \pm \sqrt{4m^2 + 9}$ and $T_2: y = mx \pm \sqrt{42m^2 - 13}$ So, $4m^2 + 9 = 42m^2 - 143$ $$\Rightarrow 38m^2 = 152$$ $$\Rightarrow m = \pm 2$$ and $c = \pm 5$ For given tangent not pass through $4^{th}$ quadrant $T: y = 2x + 5$ Now, comparing with $\frac{xx_1}{4} + \frac{yy_1}{9} = 1$ We get, $\frac{x_1}{8} = -\frac{1}{5} \Rightarrow x_1 = -\frac{8}{5}$ $$\frac{xx_2}{42} - \frac{yy_2}{143} = 1$$ $2x - y = -5$ we have $$x_2 = -\frac{84}{5}$$

Question 30

Maths · Vector Algebra · Numerical

Let $\vec{a}$, $\vec{b}$, $\vec{c}$ be three non-coplanar vectors such that $\vec{a}$ $\times$ $\vec{b}$ = 4 $\vec{c}$, $\vec{b}$ $\times$ $\vec{c}$ = 9 $\vec{a}$ and $\vec{c}$ $\times$ $\vec{a}$ = $\alpha$ $\vec{b}$, $\alpha$ > 0. If | $\vec{a}$ | + | $\vec{b}$ | + | $\vec{c}$ | = $\frac{1}{36}$, then $\alpha$ is equal to .

Answer: 36

Solution

Given $\vec{a} \times \vec{b} = 4 \vec{c}$, $\Rightarrow \vec{a} \cdot \vec{c} = 0 = \vec{b} \cdot \vec{c}$. $\vec{b} \times \vec{c} = 9 \vec{a}$, $\Rightarrow \vec{a} \cdot \vec{b} = 0 = \vec{a} \cdot \vec{c}$. Therefore, $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular set of vectors. $\Rightarrow |\vec{a}||\vec{b}| = 4|\vec{c}|$, $|\vec{b}||\vec{c}| = 9|\vec{a}|$ and $|\vec{c}||\vec{a}| = \alpha |\vec{b}|$. $\Rightarrow \frac{|\vec{a}|}{|\vec{c}|} = \frac{4}{9} \frac{|\vec{c}|}{|\vec{a}|}$. $\Rightarrow \frac{|\vec{c}|}{|\vec{a}|} = \frac{3}{2}$. If $|\vec{a}| = \lambda$, $|\vec{c}| = \frac{3 \lambda}{2}$ and $|\vec{b}| = 6$. Now $|\vec{a}| + |\vec{b}| + |\vec{c}| = \frac{1}{36}$. $\Rightarrow \frac{5}{2} \lambda + 6 = \frac{1}{36}$, $\lambda = \frac{-43}{18} = |\vec{a}|$. which gives negative value of $\lambda$ or $|\vec{a}|$ which is not possible and hence data seems to be wrong. But if $|\vec{a}| + |\vec{b}| + |\vec{c}| = 36$ $$\frac{5}{2} \lambda + 6 = 36$$ $$\lambda = 12$$ $$\alpha = \frac{|\vec{c}||\vec{a}|}{|\vec{b}|} = \frac{3 \times 12}{2} \times \frac{12}{6}$$ $$\alpha = 36$$

Physics

Question 31

Physics · Physical World, Units and Measurements · Single correct

An expression of energy density is given by $$u = \frac{\alpha}{\beta} \sin \left( \frac{\alpha x}{kt} \right),$$ where $\alpha$, $\beta$ are constants, $x$ is displacement, $k$ is Boltzmann constant and $t$ is the temperature. The dimensions of $\beta$ will be:

  1. $[M L^2 T^{-2} \theta^{-1}]$
  2. $[M^0 L^2 T^{-2}]$
  3. $[M^0 L^0 T^0]$
  4. $[M^0 L^2 T^0]$

Answer: (d)

Solution

Given $\($ $\frac{\alpha [L]}{[ML^2 \, T^{-2}]}$ = [M^0 $\,$ L^0 $\,$ T^0] $\)$. Therefore, $\($ $\alpha$ = [ML^1 $\,$ T^{-2}] $\)$. Now, $\($ $\frac{\alpha}{\beta}$ = $\frac{[ML^2 \, T^{-2}]}{[L^3]}$ $\Rightarrow$ $\beta$ = $\frac{[ML^1 \, T^{-2}][L^3]}{ML^2 \, T^{-2}}$ $\)$.

Question 32

Physics · Motion in a Plane · Single correct

A body of mass 10 kg is projected at an angle of 45° with the horizontal. The trajectory of the body is observed to pass through a point (20, 10). If T is the time of flight, then its momentum vector, at time $t = \frac{T}{\sqrt{2}}$, is ________ [Take $g = 10 \, \mathrm{m/s^2}$]

  1. $100\hat{i} + (100\sqrt{2} - 200)\hat{j}$
  2. $100\sqrt{2} \hat{i} + (100 - 200\sqrt{2})\hat{j}$
  3. $100 \hat{i} + (100 - 200\sqrt{2})\hat{j}$
  4. $100\sqrt{2} \hat{i} + (100\sqrt{2} - 200)\hat{j}$

Answer: (d)

Solution

Given $$y = x - \frac{10x^2}{2u^2 \left( \frac{1}{2} \right)} \Rightarrow 10 = 20 - \frac{(10)(100)}{u^2}$$ Let $u = 20$. Then $$T = \frac{(2)(20)}{\sqrt{2}(10)} = 2\sqrt{2}$$ Velocity $$\vec{v} = 10\sqrt{2} \hat{i} + (10\sqrt{2} - 10(2)) \hat{j}$$ Momentum $$\vec{p} = M \vec{v} = 100\sqrt{2} \hat{i} + (100\sqrt{2} - 200) \hat{j}$$

Question 33

Physics · Laws of Motion · Single correct

A block of mass M slides down on a rough inclined plane with constant velocity. The angle made by the incline plane with horizontal is $\theta$. The magnitude of the contact force will be:

  1. $Mg$
  2. $Mg \cos \theta$
  3. $\sqrt{Mg \sin \theta + Mg \cos \theta}$
  4. $Mg \sin \theta \sqrt{1 + \mu}$

Answer: (a)

Solution

Given the forces on the inclined plane, we have: The normal force is given by $N = Mg \cos \theta$. The frictional force is $f = Mg \sin \theta$. The resultant force $R$ is calculated as: $$R = \sqrt{N^2 + f^2}$$ Since $R = Mg$, the forces are balanced.

Question 34

Physics · Laws of Motion · Single correct

A block 'A' takes 2 s to slide down a frictionless incline of $30^\circ$ and length '$l$', kept inside a lift going up with uniform velocity '$v$'. If the incline is changed to $45^\circ$, the time taken by the block, to slide down the incline, will be approximately:

  1. 2.66 $\,$ $\mathrm{s}$
  2. 0.83 $\,$ $\mathrm{s}$
  3. 1.68 $\,$ $\mathrm{s}$
  4. 0.70 $\,$ $\mathrm{s}$

Answer: (c)

Solution

Given $a = g \sin \theta$. $$\ell = \frac{1}{2} g \sin 30^\circ (2)^2 \ldots (1)$$ $$\ell = \frac{1}{2} g \sin 45^\circ \ t^2 \ldots (2)$$ $$\left( \frac{1}{2} \right) (4) = \frac{1}{\sqrt{2}} t^2 \Rightarrow t = \sqrt{2} \sqrt{2} \approx 1.68$$

Question 35

Physics · System of Particles and Rotational Motion · Single correct

The velocity of the bullet becomes one third after it penetrates 4 cm in a wooden block. Assuming that bullet is facing a constant resistance during its motion in the block. The bullet stops completely after travelling at (4 + x) cm inside the block. The value of x is:

  1. 2.0
  2. 1.0
  3. 0.5
  4. 1.5

Answer: (c)

Solution

Given $\($ $\left$( $\frac{V}{3}$ $\right$)^2 = V^2 - 2a(4) $\Rightarrow$ a = $\frac{8V^2}{9(8)}$ = $\frac{V^2}{9}$ $\)$ $\($ 0 = V^2 - 2a(4 + x) $\)$ $\($ $\Rightarrow$ V^2 = 2 $\left$( $\frac{V^2}{9}$ $\right$) (4 + x) $\)$ $\($ 4.5 = 4 + x $\)$ $\($ x = 0.5 $\)$

Question 36

Physics · Gravitation · Single correct

A body of mass $m$ is projected with velocity $\lambda v_e$ in vertically upward direction from the surface of the earth into space. It is given that $v_e$ is escape velocity and $\lambda < 1$. If air resistance is considered to be negligible, then the maximum height from the centre of earth, to which the body can go, will be (R : radius of earth)

  1. $\frac{R}{1+\lambda^2}$
  2. $\frac{R}{1-\lambda^2}$
  3. $\frac{R}{1-\lambda}$
  4. $\frac{\lambda^2 R}{1-\lambda^2}$

Answer: (b)

Solution

The equation for gravitational potential energy and kinetic energy is given by: $$ -\frac{GMm}{R} + \frac{1}{2} m \lambda^2 V_e^2 = -\frac{GMm}{h} $$ Substituting the expression for escape velocity $V_e^2 = \frac{2GM}{R}$, we have: $$ -\frac{GMm}{R} + \frac{1}{2} \lambda^2 \frac{2GMm}{R} = -\frac{GMm}{h} $$ Simplifying, we get: $$ \frac{\lambda^2}{R} - \frac{1}{R} = -\frac{1}{h} $$ Rearranging gives: $$ \frac{1}{h} = \frac{1 - \lambda^2}{R} $$ Thus, the height $h$ is: $$ h = \frac{R}{1 - \lambda^2} $$

Question 37

Physics · Mechanical Properties of Solids · Single correct

A steel wire of length 3.2 m $(Y_s = 2.0 \times 10^{11} \, \mathrm{Nm}^{-2})$ and a copper wire of length 4.4 M $(Y_c = 1.1 \times 10^{11} \, \mathrm{Nm}^{-2})$, both of radius 1.4 mm are connected end to end. When stretched by a load, the net elongation is found to be 1.4 mm. The load applied, in Newton, will be: (Given $\pi = \frac{22}{7}$)

  1. 360
  2. 180
  3. 1080
  4. 154

Answer: (d)

Solution

Given $y_{steel} = 2 \times 10^{11}$ and $y_{cm} = 1.1 \times 10^{11}$. The total change in length is given by $$\Delta \ell_1 + \Delta \ell_2 = \Delta \ell$$ $$\frac{F \ell_1}{A_1 y_1} + \frac{F \ell_2}{A_2 y_2} = \Delta \ell$$ Solving for $F$, we have $$F = \frac{\Delta \ell}{\frac{\ell_1}{A_1 y_1} + \frac{\ell_2}{A_2 y_2}} = 1.54 \times 10^2 = 154$$

Question 38

Physics · Thermodynamics · Single correct

In $1^{st}$ case, Carnot engine operates between temperatures $300 \, \mathrm{K}$ and $100 \, \mathrm{K}$. In $2^{nd}$ case, as shown in the figure, a combination of two engines is used. The efficiency of this combination (in $2^{nd}$ case) will be:

  1. same as the $1^{st}$ case
  2. always greater than the $1^{st}$ case
  3. always less than the $1^{st}$ case
  4. may increase or decrease with respect to the $1^{st}$ case

Answer: (a)

Solution

First case: $\eta = 1 - \frac{100}{300} = \frac{2}{3}$ Second case: $\eta_{net} = \eta_1 + \eta_2 - \eta_1 \eta_2$ $\eta_1 = 1 - \frac{200}{300} = \frac{1}{3}$ $\eta_2 = 1 - \frac{100}{200} = \frac{1}{2}$ $\eta_{net} = \frac{1}{3} + \frac{1}{2} - \frac{1}{6} = \frac{2}{3}$ $\eta$ (first case) = $\eta$ (second case)

Question 39

Physics · Kinetic Theory · Single correct

Which statements are correct about degrees of freedom? A. A molecule with $n$ degrees of freedom has $n^2$ different ways of storing energy. B. Each degree of freedom is associated with $\frac{1}{2}RT$ average energy per mole. C. A monoatomic gas molecule has 1 rotational degree of freedom where as diatomic molecule has 2 rotational degrees of freedom D. $\mathrm{CH}_4$ has a total to 6 degrees of freedom Choose the correct answer from the option given below:

  1. B and C only
  2. B and D only
  3. A and B only
  4. C and D only

Answer: (b)

Solution

Methane molecule is tetrahedron. Degree of freedom due to rotation = 3. Degree of freedom due to translation = 3.

Question 40

Physics · Electric Charges and Fields · Single correct

A charge of $4 \, \mu \mathrm{C}$ is to be divided into two. The distance between the two divided charges is constant. The magnitude of the divided charges so that the force between them is maximum, will be:

  1. $1\,µ\mathrm{C}$ and $3\,µ\mathrm{C}$
  2. $2\,µ\mathrm{C}$ and $2\,µ\mathrm{C}$
  3. $0\,µ\mathrm{C}$ and $4\,µ\mathrm{C}$
  4. $1.5\,µ\mathrm{C}$ and $2.5\,µ\mathrm{C}$

Answer: (b)

Solution

The force $F$ is given by the equation $$F = \frac{Kq(4-q)}{d^2}.$$ Differentiating $F$ with respect to $q$, we have $$\frac{dF}{dq} = \frac{K}{d^2}[4 - 2q] = 0.$$ Solving for $q$, we find $$q = 2.$$

Question 41

Physics · Current Electricity · Single correct

A. The drift velocity of electrons decreases with the increase in the temperature of conductor. B. The drift velocity is inversely proportional to the area of cross-section of given conductor. C. The drift velocity does not depend on the applied potential difference to the conductor. D. The drift velocity of electron is inversely proportional to the length of the conductor. E. The drift velocity increases with the increase in the temperature of conductor. Choose the correct answer from the options given below:

  1. A and B only
  2. A and D only
  3. B and E only
  4. B and C only

Answer: (b)

Solution

Drift velocity = $\left$( $\frac{e \tau}{m}$ $\right$) E $$v_d = \left( \frac{e \tau}{m} \right) \left( \frac{\Delta V}{\ell} \right)$$ $\Delta V$ = Potential difference applied across the wire. As temperature increases, relaxation time decreases, hence $v_d$ decreases. As per formula, $v_d \propto \frac{1}{\ell}$ $$v_d = \frac{I}{neA}$$ As it is not mentioned that current is at steady state neither it is mentioned that $n$ is constant for given conductor. So it can't be said that $v_d$ is inversely proportional to $A$. $I = neAv_d = \frac{V}{R} = \frac{V}{\rho \ell} A$ $$v_d = \frac{V}{\rho ne} \left( E = \frac{V}{\ell} \right)$$ $$v_d = \frac{eE \tau}{m}$$ $\tau$ decreases with temperature increase. First and fourth statements are correct.

Question 42

Physics · Magnetism and Matter · Single correct

A compass needle of oscillation magnetometer oscillates 20 times per minute at a place $P$ of dip $30^\circ$. The number of oscillations per minute become 10 at another place $Q$ of $60^\circ$ dip. The ratio of the total magnetic field at the two places $(B_Q : B_P)$ is:

  1. $\sqrt{3}$ : 4
  2. 4 : $\sqrt{3}$
  3. $\sqrt{3}$ : 2
  4. 2 : $\sqrt{3}$

Answer: (a)

Solution

Given $T = 2\pi \sqrt{\frac{I}{B_H M}}$. For $T_1 = 3 sec = 2\pi \sqrt{\frac{I}{(B_P \cos 30^\circ) M}}$. For $T_2 = 6 sec = 2\pi \sqrt{\frac{I}{(B_Q \cos 60^\circ) M}}$. $$\frac{3}{6} = \sqrt{\frac{1}{B_P \left(\frac{\sqrt{3}}{2}\right)}} \times \left(\frac{B_Q}{2}\right)$$ $$\frac{3}{6} = \sqrt{\frac{B_Q}{\sqrt{3} B_P}}$$ $$\frac{\sqrt{3}}{4} = \frac{B_Q}{B_P}$$ Therefore, $B_Q : B_P = \sqrt{3} : 4$.

Question 43

Physics · Moving Charges and Magnetism · Single correct

A cyclotron is used to accelerate protons. If the operating magnetic field is 1.0 T and the radius of the cyclotron 'dees' is 60 cm, the kinetic energy of the accelerated protons in MeV will be: [use $m_p = 1.6 \times 10^{-27} \, \mathrm{kg}$, $e = 1.6 \times 10^{-19} \, \mathrm{C}$]

  1. 12
  2. 18
  3. 16
  4. 32

Answer: (b)

Solution

Kinetic energy of electron in cyclotron $$= \left[ \frac{q^2 B^2 r_0^2}{2m} \right]$$ $$= 18 \, \mathrm{MeV}$$

Question 44

Physics · Alternating Current · Single correct

A series LCR circuit has $L = 0.01 \, \mathrm{H}$, $R = 10 \, \Omega$ and $C = 1 \, \mu \mathrm{F}$ and it is connected to ac voltage of amplitude $(V_m) \, 50 \, \mathrm{V}$. At frequency 60$\%$ lower than resonant frequency, the amplitude of current will be approximately:

  1. 466 \, $\mathrm{mA}$
  2. 312 \, $\mathrm{mA}$
  3. 238 \, $\mathrm{mA}$
  4. 196 \, $\mathrm{mA}$

Answer: (c)

Solution

Resonant frequency, $\omega_0 = \frac{1}{\sqrt{LC}} = 10^4 \, rad/sec$. $\omega' = 0.4 \times 10^4 = 4000 \, rad/sec$. $i_0 = \frac{V_0}{\sqrt{R^2 + (X'_C - X'_L)^2}} = 238 \, mA$.

Question 45

Physics · Electromagnetic Waves · Single correct

Identify the correct statements from the following descriptions of various properties of electromagnetic waves. A. In a plane electromagnetic wave electric field and magnetic field must be perpendicular to each other and direction of propagation of wave should be along electric field or magnetic field. B. The energy in electromagnetic wave is divided equally between electric and magnetic fields. C. Both electric field and magnetic field are parallel to each other and perpendicular to the direction of propagation of wave. D. The electric field, magnetic field and direction of propagation of wave must be perpendicular to each other. E. The ratio of amplitude of magnetic field to the amplitude of electric field is equal to speed of light. Choose the most appropriate answer from the options given below:

  1. D only
  2. B and D only
  3. B, C and E only
  4. A, B and E only

Answer: (b)

Solution

Second and fourth statements are correct.

Question 46

Physics · Wave Optics · Single correct

Two coherent sources of light interfere. The intensity ratio of two sources is 1 : 4. For this interference pattern if the value of $\frac{I_{max} + I_{min}}{I_{max} - I_{min}}$ is equal to $\frac{2\alpha + 1}{\beta + 3}$, then $\frac{\alpha}{\beta}$ will be:

  1. 1.5
  2. 2
  3. 0.5
  4. 1

Answer: (b)

Solution

Given $\frac{I_1}{I_2} = \frac{1}{4}$. Therefore, $I_2 = 4I_1$. The maximum current $I_{max} = I_1 + 4I_1 + 2\sqrt{I_1 \cdot 4I_1} = 9I_1$. The minimum current $I_{min} = I_1 + 4I_1 - 2\sqrt{I_1 \cdot 4I_1} = I_1$. Thus, $\frac{9I_1 + I_1}{9I_1 - I_1} = \frac{10}{8} = \frac{5}{4} = \frac{2\alpha + 1}{\beta + 1}$. Solving gives $\alpha = 2$ and $\beta = 1$. Therefore, $\frac{\alpha}{\beta} = \frac{2}{1} = 2$.

Question 47

Physics · Dual Nature of Radiation and Matter · Single correct

With reference to the observations in photo-electric effect, identify the correct statements from below: A. The square of maximum velocity of photoelectrons varies linearly with frequency of incident light. B. The value of saturation current increases on moving the source of light away from the metal surface. C. The maximum kinetic energy of photo-electrons decreases on decreasing the power of LED (light emitting diode) source of light. D. The immediate emission of photo-electrons out of metal surface can not be explained by particle nature of light/electromagnetic waves. E. Existence of threshold wavelength can not be explained by wave nature of light/electromagnetic waves. Choose the correct answer from the options given below:

  1. A and B only
  2. A and E only
  3. C and E only
  4. D and E only

Answer: (b)

Solution

The equation for the photoelectric effect is given by $$\frac{1}{2} m V_{max}^2 = h f - \phi$$ where $m$ is the mass, $V_{max}$ is the maximum velocity, $h$ is Planck's constant, $f$ is the frequency, and $\phi$ is the work function. The photoelectric effect can be explained by the particle nature of light. The threshold $\lambda$ is the maximum wavelength at which emission takes place.

Question 48

Physics · Nuclei · Single correct

The activity of a radioactive material is $6.4 \times 10^{-4}$ curie. Its half life is 5 days. The activity will become $5 \times 10^{-6}$ curie after:

  1. 7 days
  2. 15 days
  3. 25 days
  4. 35 days

Answer: (d)

Solution

Given $A_0 = 6.4 \times 10^{-4} Curie$. The half-life $T_{1/2} = 5 days = \frac{\ln 2}{\lambda}$. The activity $A = A_0 e^{-\lambda t}$. $$5 \times 10^{-6} = 6.4 \times 10^{-4} e^{-\lambda t}$$ $$\frac{5}{6.4} \times 10^{-2} = e^{-\lambda t}$$ $$7.8 \times 10^{-3} = e^{-\lambda t}$$ $$\log(7.8 \times 10^{-3}) = -\lambda t \ln e$$ $$\ln(7.8 \times 10^{-3}) = -\frac{\lambda \ln 2}{5} \cdot t$$ $$5 \times \frac{4.853}{0.693} = t = 35 days$$

Question 49

Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct

For a constant collector-emitter voltage of $8\,\mathrm{V}$, the collector current of a transistor reached to the value of $6\,\mathrm{mA}$ from $4\,\mathrm{mA}$, whereas base current changed from $20\,\mathrm{µA}$ to $25\,\mathrm{µA}$ value. If transistor is in active state, small signal current gain (current amplification factor) will be:

  1. 240
  2. 400
  3. 0.0025
  4. 200

Answer: (b)

Solution

Given $V_{\mathrm{CE}} = 8 \, \mathrm{V}$, $I_{\mathrm{C}} = 6 \, \mathrm{mA}$ from $4 \, \mathrm{mA}$, $I_{\mathrm{B}} = 20 \, \mu\mathrm{A}$ to $25 \, \mu\mathrm{A}$. Current gain $\beta_{\mathrm{av}}$ is calculated as follows: $$\beta_{\mathrm{av}} = \frac{I_{\mathrm{C}}' - I_{\mathrm{C}}}{I_{\mathrm{B}}' - I_{\mathrm{B}}} = \frac{2 \, \mathrm{mA}}{5 \, \mu\mathrm{A}}$$ $$\beta_{\mathrm{av}} = \frac{2}{5} \times 10^3 = \frac{2000}{5} = 400$$

Question 50

Physics · Communication Systems · Single correct

A square wave of the modulating signal is shown in the figure. The carrier wave is given by $C(t) = 5 \sin (8 \pi t)$ Volt. The modulation index is:

  1. 0.2
  2. 0.1
  3. 0.3
  4. 0.4

Answer: (a)

Solution

Modulation Index $\mu = \frac{A_m}{A_C} = \frac{1}{5} = 0.2$ $A_m =$ amp. of modulating signal $A_C =$ amp. of carrier wave

Question 51

Physics · Mechanical Properties of Solids · Numerical

In an experiment to determine the Young's modulus, steel wires of five different lengths (1, 2, 3, 4 and 5 m) but of same cross section ($2\,\mathrm{mm}^2$) were taken and curves between extension and load were obtained. The slope (extension/load) of the curves were plotted with the wire length and the following graph is obtained. If the Young's modulus of given steel wires is $x \times 10^{11} \, \mathrm{Nm}^{-2}$, then the value of $x$ is ________ .

Answer: 2

Solution

Slope = $\frac{\Delta l / w}{L}$ = $\frac{\Delta l / L}{w}$ = $\frac{1}{YA}$ $\Rightarrow$ Y = $\frac{1}{(slope)A}$ Y = $\frac{1}{2 \times 10^{-6} (0.25 \times 10^{-5})}$ Y = 2 $\times$ $10^{11}$ $\mathrm{N/m^2}$

Question 52

Physics · Current Electricity · Numerical

In the given figure of meter bridge experiment, the balancing length AC corresponding to null deflection of the galvanometer is 40 cm. The balancing length, if the radius of the wire AB is doubled, will be……….cm.

Answer: 40

Solution

Independent of area in case of uniform wire.

Question 53

Physics · Ray Optics and Optical Instruments · Numerical

A thin prism of angle $6^\circ$ and refractive index for yellow light $(n_Y)1.5$ is combined with another prism of angle $5^\circ$ and $n_Y = 1.55$. The combination produces no dispersion. The net average deviation $(\delta)$ produced by the combination is $\left( \frac{1}{x} \right)^\circ$. The value of $x$ is.......

Answer: 4

Solution

Given $\delta = A(\mu_y - 1) - A'(\mu_y' - 1)$. $$= 6(1.5 - 1) - 5(1.55 - 1)$$ $$= \frac{1}{4}$$

Question 54

Physics · Electromagnetic Induction · Numerical

A conducting circular loop is placed in X - Y plane in presence of magnetic field $\vec{B} = (3t^3 \hat{j} + 3t^2 \hat{k})$ in SI unit. If the radius of the loop is $1 \, \mathrm{m}$, the induced emf in the loop, at time, $t = 2 \, \mathrm{s}$ is $n \pi \, \mathrm{V}$. The value of $n$ is........

Answer: 12

Solution

The magnetic flux $\phi$ is given by $\phi = \mathbf{B} \cdot \mathbf{A}$. This is equal to $\phi = (3t^3 \hat{\mathbf{j}} + 3t^2 \hat{\mathbf{k}}) \cdot (\pi (1)^2 \hat{\mathbf{k}})$. Simplifying, we get $\phi = 3t^2 \pi$. The induced emf $\varepsilon_{IND}$ is given by $\varepsilon_{IND} = \left| \frac{d\phi}{dt} \right| = 6t\pi$. At $t = 2$, $\varepsilon_{IND} = 12$.

Question 55

Physics · Electrostatic Potential and Capacitance · Numerical

As shown in the figure, in steady state, the charge stored in the capacitor is....... $\times 10^{-6} \mathrm{C}$.

Answer: 10

Solution

Given the equation for charge, we have: $$q = C V_{100\Omega}$$ Substituting the values, we get: $$= (1.1 \times 10^{-6}) \left( \frac{10}{R + r} R \right)$$ Simplifying further: $$= 1.1 \times 10^{-6} \left( \frac{10}{110} \times 100 \right)$$ Finally, we find: $$= 10 \, \mu C$$

Question 56

Physics · Electrostatic Potential and Capacitance · Fill in the blank

A parallel plate capacitor with width 4 cm, length 8 cm and separation between the plates of 4 $\mathrm{mm}$ is connected to a battery of 20 $\mathrm{V}$. A dielectric slab of dielectric constant 5 having length 1 $\mathrm{cm}$, width 4 $\mathrm{cm}$ and thickness 4 $\mathrm{mm}$ is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be ........ $\varepsilon_0$ $\mathrm{J}$. (Where $\varepsilon_0$ is the permittivity of free space)

Answer: 240

Solution

The effective capacitance is given by $$C_{eff} = \left[ \frac{\varepsilon_0 (7 \times 4)}{4/10} + \frac{5 \varepsilon_0 (1 \times 4)}{4/10} \right] \times 10^{-2}$$ Simplifying, we find $$C_{eff} = 1.2 \varepsilon_0$$ The energy is calculated as $$Energy = \frac{1}{2} C_{eff} V^2$$ Substituting the values, we get $$= \frac{1}{2} (1.2) \varepsilon_0 (20)(20) = 240 \varepsilon_0$$

Question 57

Physics · Waves · Numerical

A wire of length 30 cm, stretched between rigid supports, has it's $n^{th}$ and $(n + 1)^{th}$ harmonics at 400 Hz and 450 Hz, respectively. If tension in the string is 2700 N, it's linear mass density is........kg/m.

Answer: 3

Solution

$\dfrac{nv}{0.6} = 400 \ \& \ \dfrac{(n+1)v}{0.6} = 450$ $\Rightarrow \left[\dfrac{0.6 \times 400}{v} + 1\right]\dfrac{v}{0.6} = 450$ $\Rightarrow v = 30$ $\Rightarrow \sqrt{\dfrac{T}{\mu}} = 30$ $\Rightarrow \dfrac{2700}{\mu} = 900 \Rightarrow \mu = 3$

Question 58

Physics · Mechanical Properties of Fluids · Numerical

A spherical soap bubble of radius 3 cm is formed inside another spherical soap bubble of radius 6 cm. If the internal pressure of the smaller bubble of radius 3 cm in the above system is equal to the internal pressure of the another single soap bubble of radius $r$ cm. The value of $r$ is.......

Answer: 2

Solution

Given $P_2 - P_0 = \frac{4T}{6}$ and $P_1 - P_2 = \frac{4T}{3}$. Therefore, $$P_1 - P_0 = \frac{4T}{2} = 2$$

Question 59

Physics · Work, Energy and Power · Numerical

A solid cylinder length is suspended symmetrically through two massless strings, as shown in the figure. The distance from the initial rest position, the cylinder should by unbinding the strings to achieve a speed of $4 \, \mathrm{ms^{-1}}$, is.......cm. (take $g = 10 \, \mathrm{ms^{-2}}$)

Answer: 120

Solution

From energy conservation $$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$ $$mgh = \frac{1}{2}mv^2 + \frac{1}{2} \frac{mR^2}{2} \omega^2$$ $$10h = \frac{16}{2} + \frac{16}{4} \Rightarrow h = 1.2\, \mathrm{m} = 120\, \mathrm{cm}$$

Question 60

Physics · Motion in a Plane · Numerical

Two inclined planes are placed as shown in figure. A block is projected from the Point A of inclined plane AB along its surface with a velocity just sufficient to carry it to the top Point B at a height 10 $\,$ $\mathrm{m}$. After reaching the Point B the block slides down on inclined plane BC. Time it takes to reach to the point C from point A is $t(\sqrt{2} + 1) \, \mathrm{s}$. The value of $t$ is........(use $g = 10 \, \mathrm{m/s^2}$)

Answer: 2

Solution

From E.C. = $\frac{1}{2} mv_0^2 = mgh$ $v_0 = 10\sqrt{2}$ For $A \to B$ at $B$, $v = 0$ $a = -g \sin 45^\circ = \frac{-10}{\sqrt{2}}$ $v = u + at_1 \Rightarrow 0 = 10\sqrt{2} - \frac{10}{\sqrt{2}} t_1 \Rightarrow t_1 = 2 sec$ For $B \to C$ $s = ut_2 + \frac{1}{2} at_2^2$ $$\frac{10}{\sin 30^\circ} = \frac{1}{2} (10 \sin 30^\circ) t_2^2$$ $t_2 = 2\sqrt{2}$ So total time $T = t_1 + t_2$ $= 2\sqrt{2} + 2$ $= 2(\sqrt{2} + 1) sec$

Chemistry

Question 61

Chemistry · Structure of Atom · Single correct

The correct decreasing order of energy for the orbitals having the following set of quantum numbers: (A) $n=3,\ l=0,\ m=0$ (B) $n=4,\ l=0,\ m=0$ (C) $n=3,\ l=1,\ m=0$ (D) $n=3,\ l=2,\ m=1$

  1. $(D) > (B) > (C) > (A)$
  2. $(B) > (D) > (C) > (A)$
  3. $(C) > (B) > (D) > (A)$
  4. $(B) > (C) > (D) > (A)$

Answer: (a)

Solution

\text{(A)}\quad $n+\ell$=3+0=3 \text{(B)}\quad $n+\ell$=4+0=4 \text{(C)}\quad $n+\ell$=3+1=4 \text{(D)}\quad $n+\ell$=3+2=5 Higher $n+\ell$ value, higher the energy \& if same $n+\ell$ value, then higher $n$ value, higher the energy. Thus : $D>B>C>A$.

Question 62

Chemistry · Chemical Bonding and Molecular Structure · Single correct

Match List-I with List-II List-I \begin{tabular}{|c|c|c|c|} \hline \multicolumn{2}{|c|}{List-I} & \multicolumn{2}{c|}{List-II} \\ \hline (A) & $\Psi_{MO}=\Psi_A-\Psi_B$ & (I) & Dipole moment \\ \hline (B) & $\mu=Q\times r$ & (II) & Bonding molecular orbital \\ \hline (C) & $\dfrac{N_b-N_a}{2}$ & (III) & Anti-bonding molecular orbital \\ \hline (D) & $\Psi_{MO}=\Psi_A+\Psi_B$ & (IV) & Bond order \\ \hline \end{tabular}

  1. (A) -(II), (B)-(I), $(C)$-(IV), (D)-(III)
  2. (A) -(III), (B)-(IV), $(C)$-(I), (D)-(II)
  3. (A) -(III), (B)-(I), $(C)$-(IV), (D)-(II)
  4. (A) -(III), (B)-(IV), $(C)$-(II), (D)-(I)

Answer: (c)

Solution

(A) $\psi_{MO} = \psi_A - \psi_B$ (B) $\mu = Q \times r$ $(C)$ $\frac{N_b - N_a}{2}$ (D) $\psi_{MO} = \psi_A + \psi_B$ (III) ABMO (I) Dipole moment (IV) Bond order (II) BMO

Question 63

Chemistry · Equilibrium · Single correct

The Plot of pH-metric titration of weak base $\mathrm{NH_4OH}$ vs strong acid $\mathrm{HCl}$ looks like:

Answer: (a)

Solution

Titration curve of $\mathrm{NH_4OH}$ vs $\mathrm{HCl}$ (WB + SA).

Question 64

Chemistry · Electrochemistry · Single correct

Given below are two statements: Statement I: For KI, molar conductivity increases steeply with dilution. Statement II: For carbonic acid, molar conductivity increases slowly with dilution. In the light of the above statements, choose the correct answer from the options given below:

  1. Both Statement I and Statement II are true
  2. Both Statement I and Statement II are false
  3. Statement I is true but Statement II is false
  4. Statement I is false but Statement II is true

Answer: (b)

Solution

Statement I: KI is a strong electrolyte thus almost constant on dilution. Statement II: In weak electrolyte it increases sharply.

Question 65

Chemistry · Surface Chemistry · Single correct

Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$ Assertion $(A)$ : Dissolved substances can be removed from a colloidal solution by diffusion through a parchment paper. Reason $(R)$ : Particles in a true solution cannot pass through parchment paper but the colloidal particles can pass through the parchment paper. In the light of the above statements, choose the correct answer from the options given below:

  1. Both (A) and $(R)$ are correct and $(R)$ is the correct explanation of (A)
  2. Both (A) and $(R)$ are correct but $(R)$ is not the correct explanation of (A)
  3. $(A)$ is correct but $(R)$ is not correct
  4. $(A)$ is not correct but $(R)$ is correct

Answer: (c)

Solution

Assertion (A): Correct. Reason (R): Incorrect. Particles of true solution pass through parchment paper thus answer is (C).

Question 66

Chemistry · Classification of Elements and Periodicity in Properties · Single correct

The outermost electronic configurations of four elements A, B, C, D are given below: (A) $3s^2$ (B) $3s^2 3p^1$ (C) $3s^2 3p^3$ (D) $3s^2 3p^4$ The correct order of first ionization enthalpy for them is:

  1. < (B) < $(C)$ < (D)
  2. < (A) < (D) < $(C)$
  3. < (D) < (A) < $(C)$
  4. < (A) < $(C)$ < (D)

Answer: (b)

Solution

Given the options: (A) $3s^2 \rightarrow Mg$ (B) $3s^2 3p^1 \rightarrow Al$ (C) $3s^2 3p^3 \rightarrow P$ (D) $3s^2 3p^4 \rightarrow S$ We have: Therefore, the order is $C > D > A > B$.

Question 67

Chemistry · The s-Block Elements · Single correct

An element A of group 1 shows similarity to an element B belonging to group 2. If A has maximum hydration enthalpy in group 1 then B is:

  1. Mg
  2. Be
  3. Ca
  4. Sr

Answer: (a)

Solution

Diagonal relationship $\mathrm{Li^+}$ has maximum hydration enthalpy in group 1 due to small size. So 'B' is $\mathrm{Mg}$.

Question 68

Chemistry · The p-Block Elements (Group-13 and 14) · Single correct

Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$ Assertion $(A)$ : Boron is unable to form $\mathrm{BF}_6^{3-}$ Reason $(R)$ : Size of B is very small. In the light of the above statements, choose the correct answer from the options given below:

  1. Both (A) and $(R)$ are true and $(R)$ is the correct explanation of (A)
  2. Both (A) and $(R)$ are true but $(R)$ is not the correct explanation of (A)
  3. is true but $(R)$ is false
  4. is false but $(R)$ is true

Answer: (b)

Solution

Assertion (A): True Reason $(R)$: True but not correct explanation. Correct explanation: Expansion of octet not possible for 'B'.

Question 69

Chemistry · The d-and f-Block Elements · Single correct

In neutral or alkaline solution, $\mathrm{MnO}_4^-$ oxidises thiosulphate to:

  1. $\mathrm{S}_2\mathrm{O}_7^{2-}$
  2. $\mathrm{S}_2\mathrm{O}_8^{2-}$
  3. $\mathrm{SO}_3^{2-}$
  4. $\mathrm{SO}_4^{2-}$

Answer: (d)

Solution

The reaction is given by: $$8\mathrm{MnO_4^-} + 3\mathrm{S_2O_3^{2-}} + \mathrm{H_2O} \xrightarrow{neutral or alk. solution} 8\mathrm{MnO_2} + 6\mathrm{SO_4^{2-}} + 2\mathrm{OH^-}$$

Question 70

Chemistry · Co-ordination Compounds · Single correct

Low oxidation state of metals in their complexes are common when ligands:

  1. have good $\pi$-accepting character
  2. have good $\sigma$-donor character
  3. are having good $\pi$-donating ability
  4. are having poor $\sigma$-donating ability

Answer: (a)

Solution

When metal is in low oxidation state then it forms complexes when ligands have good $\pi$-accepting character.

Question 71

Chemistry · Environmental Chemistry · Single correct

Given below are two statements: Statement I : The non bio-degradable fly ash and slag from steel industry can be used by cement industry. Statement II : The fuel obtained from plastic waste is lead free. In the light of the above statements, choose the most appropriate answer from the options given below:

  1. Both Statement I and Statement II are correct
  2. Both Statement I and Statement II are incorrect
  3. Statement I is correct but Statement II is incorrect
  4. Statement I is incorrect but Statement II is correct

Answer: (a)

Solution

(I) Fly ash and slag from steel industry are utilised by cement industry. (II) Fuel obtained from plastic waste has high octane rating. It contains no lead and it is known as green fuel. Both statement (I) & (II) are correct.

Question 72

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

The structure of A in the given reaction is:

Answer: (c)

Solution

The reaction involves the deprotonation of the ketone by $\mathrm{NaOH}$ to form an enolate ion. This enolate ion then undergoes an alkylation reaction with ethyl bromide to form the final product.

Question 73

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

Major product 'B' of the following reaction sequence is:

Answer: (b)

Solution

The reaction starts with the compound $\mathrm{CH_3-C=CH-CH_3}$ with a $\mathrm{CH_3}$ group attached to the second carbon. It reacts with $\mathrm{Br_2}$ in $\mathrm{CH_3OH}$ to form $\mathrm{CH_3-CH(OCH_3)-CHBr-CH_3}$. This intermediate then reacts with $\mathrm{HI}$ to form the final product $\mathrm{CH_3-CH(I)-CHBr-CH_3}$.

Question 74

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

Match List-I with List-II.\ List-II\ (I) Gatterman Koch reaction\ (II) Etard reaction\ (III) Stephen reaction\ (IV) Rosenmund reaction\ Choose the correct answer from the options given below:

  1. (A) -(IV), (B)-(III), $(C)$-(II), (D)-(I)
  2. (A)-(I), (B)-(II), $(C)$-(III), (D)-(IV)
  3. (A)-(II), (B)-(III), $(C)$-(IV), (D)-(I)
  4. (A)-(III), (B)-(II), $(C)$-(I), (D)-(IV)

Answer: (a)

Solution

Option (A) involves the conversion of benzoyl chloride to benzaldehyde using hydrogen gas and palladium on barium sulfate as a catalyst. This is known as the Rosenmund reaction. Option (B) involves the conversion of acetonitrile to acetaldehyde using tin(II) chloride and hydrochloric acid, followed by hydrolysis. This is known as the Stephen reaction. Option (C) involves the oxidation of toluene to benzaldehyde using chromyl chloride and acid. This is known as the Etard reaction. Option (D) involves the formylation of benzene to benzaldehyde using carbon monoxide and hydrochloric acid in the presence of anhydrous aluminum chloride. This is known as the Gattermann Koch reaction.

Question 75

Chemistry · Polymers · Single correct

Match List-I with List-II. \begin{tabular}{|c|l|c|l|} \hline \multicolumn{2}{|c|}{List-I} & \multicolumn{2}{c|}{List-II} \\ \multicolumn{2}{|c|}{(Polymer)} & \multicolumn{2}{c|}{(Monomer)} \\ \hline (A) & Neoprene & (I) & Acrylonitrile \\ \hline (B) & Teflon & (II) & Chloroprene \\ \hline (C) & Acrilan & (III) & Tetrafluoroethene \\ \hline (D) & Natural rubber & (IV) & Isoprene \\ \hline \end{tabular} Choose the correct answer from the option given below:

  1. (A) -(II), (B)-(III), $(C)$-(I), (D-(IV)
  2. (A) -(II), (B)-(I), $(C)$-(III), (D-(IV)
  3. (A) -(II), (B)-(I), $(C)$-(IV), (D-(III)
  4. (A) -(I), (B)-(II), $(C)$-(III), (D-(IV)

Answer: (a)

Solution

Question 76

Chemistry · Amines · Single correct

An organic compound ‘A’ contains nitrogen and chlorine. It dissolves readily in water to give a solution that turns litmus red. Titration of compound ‘A’ with standard base indicates that the molecular weight of ‘A’ is $131 \pm 2$. When a sample of ‘A’ is treated with aq. NaOH, a liquid separates which contains N but not Cl. Treatment of the obtained liquid with nitrous acid followed by phenol gives orange precipitate. The compound ‘A’ is :

Answer: (d)

Solution

Question 77

Chemistry · Biomolecules · Single correct

Match List-I with List-II.\ List-I\ (A) Glucose + HI\ (B) Glucose + $Br_2$ water\ (C) Glucose + acetic anhydride\ (D) Glucose + $HNO_3$\ List-II\ (I) Gluconic acid\ (II) Glucose pentacetate\ (III) Saccharic acid\ (IV) Hexane\ Choose the correct answer from the options given below:

  1. (A) -(IV), (B)-(I), $(C)$-(II), (D)-(III)
  2. (A)-(IV), (B)-(III), $(C)$-(II), (D)-(I)
  3. (A)-(III), (B)-(I), $(C)$-(IV), (D)-(II)
  4. (A)-(I), (B)-(III), $(C)$-(IV), (D)-(II)

Answer: (a)

Solution

Option (A) shows the reaction of glucose with HI to form n-hexane. Option (B) shows the reaction of glucose with $\mathrm{Br_2}$ and $\mathrm{H_2O}$ to form gluconic acid. Option $(C)$ shows the reaction of glucose with 5 acetic anhydride to form glucose pentacetate. Option (D) shows the reaction of glucose with $\mathrm{HNO_3}$ to form saccharic acid.

Question 78

Chemistry · Chemistry in Everyday Life · Single correct

Which of the following enhances the lathering property of soap?

  1. Sodium stearate
  2. Sodium carbonate
  3. Sodium rosinate
  4. Trisodium phosphate

Answer: (c)

Solution

Rosin is added to soaps which forms sodium rosinate which lathers well.

Question 79

Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct

Match List-I with List-II \begin{tabular}{|l|l|} \hline \textbf{List-I} & \textbf{List-II} \\ (Mixture) & (Purification Process) \\ \hline (A) Chloroform & (I) Steam distillation \\ \& Aniline & \\ \hline (B) Benzoic acid & (II) Sublimation \\ \& Naphthalene & \\ \hline (C) Water & (III) Distillation \\ \& Aniline & \\ \hline (D) Naphthalene & (IV) Crystallisation \\ \& Sodium chloride & \\ \hline \end{tabular}

  1. (A) -(IV), (B)-(III), ($C$)-(I), (D)-(II)
  2. (A) -(III), (B)-(I), ($C$)-(IV), (D)-(II)
  3. (A) -(III), (B)-(IV), ($C$)-(II), (D)-(I)
  4. (A) -(III), (B)-(IV), ($C$)-(I), (D)-(II)

Answer: (d)

Solution

(A) Chloroform + Aniline $\rightarrow$ ($\mathrm{III}$) Distillation (B) Benzoic acid + Napthalene $\rightarrow$ ($\mathrm{IV}$) Crystallisation ($C$) Water + Aniline $\rightarrow$ ($\mathrm{I}$) Steam distillation (D) Napthalene + Sodium chloride $\rightarrow$ ($\mathrm{II}$) Sublimation

Question 80

Chemistry · Co-ordination Compounds · Single correct

$Fe^{3+}$ cation gives a prussian blue precipitate on addition of potassium ferrocyanide solution due to the formation of:

  1. [$\mathrm{Fe(H_2O)_6}]_2$ $[\mathrm{Fe(CN)_6}$]
  2. $\mathrm{Fe_2[Fe(CN)_6]_2}$
  3. $\mathrm{Fe_3[Fe(OH)_2(CN)_4]_2}$
  4. $\mathrm{Fe_4[Fe(CN)_6]_3}$

Answer: (d)

Solution

The reaction is given by: $$4 \mathrm{Fe}^{3+} + 3[\mathrm{Fe(CN)}_6]^{-4} \rightarrow \mathrm{Fe}_4[\mathrm{Fe(CN)}_6]_3$$ This forms Prussian Blue.

Question 81

Chemistry · Some Basic Concepts of Chemistry · Numerical

The normality of $\mathrm{H_2SO_4}$ in the solution obtained on mixing $100 \, \mathrm{mL}$ of $0.1 \, \mathrm{M} \, \mathrm{H_2SO_4}$ with $50 \, \mathrm{mL}$ of $0.1 \, \mathrm{M} \, \mathrm{NaOH}$ is _______ $\times 10^{-1} \, \mathrm{N}$. (Nearest Integer)

Answer: 1

Solution

No. of equivalents of $\mathrm{H_2SO_4} = 100 \times 0.1 \times 2 = 20$ No. of equivalents of $\mathrm{NaOH} = 50 \times 0.1 = 5$ No. of equivalents of $\mathrm{H_2SO_4}$ left $= 20 - 5 = 15$ $$150 \times x = 15$$ $$x = \frac{1}{10} = 0.1\, \mathrm{N} = 1 \times 10^{-1}\, \mathrm{N}$$

Question 82

Chemistry · States of Matter · Numerical

for a real gas at $25^\circ \mathrm{C}$ temperature and high pressure ($99 \mathrm{\ bar}$) the value of compressibility factor is 2, so the value of Vander Waal’s constant ‘b’ should be _____ $\times 10^{-2} \mathrm{\ L \ mol^{-1}}$ (Nearest integer) (Given $R = 0.083 \mathrm{L \ bar \ K}^{-1} \mathrm{mol}^{-1}$)

Answer: 25

Solution

For real gas under high pressure, $$Z = 1 + \frac{Pb}{RT}$$ implies $$b = \frac{RT}{P}$$ $$= \frac{0.083 \times 298}{99}$$ $$= 0.25 \times 10^{-2} \, \mathrm{L \, mol^{-1}}$$

Question 83

Chemistry · Thermodynamics · Numerical

A gas (Molar mass = 280 $\mathrm{g \, mol^{-1}}$) was burnt in excess $\mathrm{O_2}$ in a constant volume calorimeter and during combustion the temperature of calorimeter increased from 298.0 $\mathrm{K}$ to 298.45 $\mathrm{K}$. If the heat capacity of calorimeter is 2.5 $\mathrm{kJ \, K^{-1}}$ and enthalpy of combustion of gas is 9 $\mathrm{kJ \, mol^{-1}}$ then amount of gas burnt is $\mathrm{g}$. (Nearest Integer)

Answer: 35

Solution

Let $x \, \mathrm{g}$ is burnt. moles = $\frac{x}{280}$ heat released by $\frac{x}{280}$ mole = 2.5 $\times$ 0.45 $\,$ $\mathrm{kJ}$ heat released by 1 mole = $\frac{2.5 \times 0.45 \times 280}{x}$ $\,$ $\mathrm{kJ}$ $\Delta$ H = $\Delta$ U + $\Delta$ n g R T $\Delta$ H $\approx$ $\Delta$ U 9 = $\frac{2.5 \times 280 \times 0.45}{x}$ x = 35 $\,$ $\mathrm{g}$

Question 84

Chemistry · Solutions · Numerical

When a certain amount of solid A is dissolved in $100\,\mathrm{g}$ of water at $25°\mathrm{C}$ to make a dilute solution, the vapour pressure of the solution is reduced to one-half of that of pure water. The vapour pressure of pure water is $23.76\,\mathrm{mmHg}$. The number of moles of solute A added is \_\_\_\_. (Nearest Integer) Assume moles of A to be less than moles of B.

Answer: 3

Solution

Dilute solution given: $$\frac{P^0 - P_s}{P^0} \sim \frac{n_{solute}}{n_{solvent}}$$ $$\frac{P^0 - \frac{P^0}{2}}{P^0} = \frac{n_{solute}}{n_{solvent}}$$ $$n_{solute} \sim \frac{n_{solvent}}{2} = \frac{100}{18 \times 2} = 2.78 \, mol$$ More accurate approach: $$\frac{P^0 - P_s}{P_s} = \frac{n_{solute}}{n_{solvent}}$$ $$\frac{P^0 - \frac{P^0}{2}}{\frac{P^0}{2}} = \frac{n_{solute}}{n_{solvent}}$$ $$n_{solute} = n_{solvent} = \frac{100}{18} = 5.55 \, mol$$

Question 85

Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical

$$ \begin{array}{ccc} [A] & \rightarrow & [B] \\ \text{Reactant} & & \text{Product} \end{array} $$ If formation of compound [B] follows the first order of kinetics and after 70 minutes the concentration of [A] was found to be half of its initial concentration. Then the rate constant of the reaction is $x\times10^{-6}\,\mathrm{s}^{-1}$. The value of $x$ is ______. (Nearest Integer)

Answer: 165

Solution

K is calculated as follows: $$K = \frac{0.693}{t_{1/2}} = \frac{0.693}{70 \times 60}$$ This simplifies to: $$= \frac{6930}{7 \times 6} \times 10^{-6}$$ Finally, we have: $$= 165 \times 10^{-6} \, \mathrm{s^{-1}}$$

Question 86

Chemistry · General Principles and Processes of Isolation of Elements · Numerical

Among the following ores Bauxite, Siderite, Cuprite, Calamine, Haematite, Kaolinite, Malachite, Magnetite, Sphalerite, Limonite, Cryolite, the number of principal ores if (of) iron is ________.

Answer: 4

Solution

Bauxite $\;-\;$ $AlO_x$ $(OH)_{3-2x}$ (where 0 < x < 1) $\checkmark$ Siderite $\;-\;$ $\mathrm{FeCO_3}$ Cuprite $\;-\;$ $\mathrm{Cu_2O}$ Calamine $\;-\;$ $\mathrm{ZnCO_3}$ $\checkmark$ Haematite $\;-\;$ $\mathrm{Fe_2O_3}$ Kaolinite $\;-\;$ $\mathrm{Al_2(OH)_4Si_2O_5}$ Malachite $\;-\;$ $\mathrm{CuCO_3\cdot Cu(OH)_2}$ $\checkmark$ Magnetite $\;-\;$ $\mathrm{Fe_3O_4}$ Sphalerite $\;-\;$ $\mathrm{ZnS}$ $\checkmark$ Limonite $\;-\;$ $\mathrm{Fe_2O_3\cdot 3H_2O}$ Cryolite $\;-\;$ $\mathrm{Na_3AlF_6}$

Question 87

Chemistry · The d-and f-Block Elements · Fill in the blank

The oxidation state of manganese in the product obtained in a reaction of potassium permanganate and hydrogen peroxide in basic medium is _______.

Answer: 4

Solution

In a basic medium, the reaction is given by: $$2 \mathrm{KMnO_4} + 3 \mathrm{H_2O_2} \xrightarrow{basic medium} 2 \mathrm{MnO_2} + 3 \mathrm{O_2} + 2 \mathrm{H_2O} + 2 \mathrm{KOH}.$$

Question 88

Chemistry · Chemical Bonding and Molecular Structure · Numerical

The number of molecule(s) or ion(s) from the following having non-planar structure is ______. $NO_3^-, H_2O_2, BF_3, PCl_3, XeF_4, SF_4, XeO_3, PH_4^+, SO_3, [Al(OH)_4]^-$

Answer: 6

Solution

$SO_3$ has $sp^2$ hybridization and is planar. $BF_3$ has $sp^2$ hybridization and is planar. $NO_3^-$ has $sp^2$ hybridization and is planar. $SF_4$ has $sp^3d$ hybridization and is non-planar. $H_2O_2$ has $sp^3$ hybridization and is non-planar. $PCl_3$ has $sp^3$ hybridization and is non-planar. $[Al(OH)_4]^-$ has $sp^3$ hybridization and is non-planar. $XeF_4$ has $sp^3d^2$ hybridization and is planar. $XeO_3$ has $sp^3$ hybridization and is non-planar. $PH_4^+$ has $sp^3$ hybridization and is non-planar.

Question 89

Chemistry · Co-ordination Compounds · Numerical

The spin only magnetic moment of the complex present in Fehling's reagent is $\dots\dots$ B.M. (Nearest integer).

Answer: 2

Solution

Fehling solution is a complex of $\mathrm{Cu^{++}}$. $$\mathrm{Cu^{++} = 3d^9}$$ Number of unpaired $e^- = 1$. $$\mathrm{M.M} = \sqrt{1(1+2)} = \sqrt{3} = 1.73 \, \mathrm{BM}$$

Question 90

Chemistry · Some Basic Concepts of Chemistry · Numerical

In the above reaction, 5 g of toluene is converted into benzaldehyde with 92% yield. The amount of benzaldehyde produced is _____ $\times$ $10^{-2}$ g. (Nearest integer)

Answer: 530

Solution

Moles $=\frac{5}{92}$ Moles of $\mathrm{CHO}=\frac{5}{92}\times\frac{92}{100}=5\times10^{-2}$ Mass of $\mathrm{CHO}=106\times5\times10^{-2}=5.3\,\mathrm{g}=530\times10^{-2}\,\mathrm{g}$