JEE Main 17 March 2021 Shift 1: 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 · Relations and Functions · Single correct
The inverse of $y = 5^{\log x}$ is:
Answer: (c)
Given $y = 5^{\log x}$. Then $y = x^{\log 5}$. Therefore, $y^{\frac{1}{\log x}} = x$. Replying $x \to y$ and $y \to x$.
Maths · Vector Algebra · Single correct
Let $\vec{a}=2\hat{i}-3\hat{j}+4\hat{k}$ and $\vec{b}=7\hat{i}+\hat{j}-6\hat{k}$. If $\vec{r}\times\vec{a}=\vec{r}\times\vec{b}$, $\vec{r}\cdot(\hat{i}+2\hat{j}+\hat{k})=-3$, then $\vec{r}\cdot(2\hat{i}-3\hat{j}+\hat{k})$ is equal to $\underline{\hspace{2cm}}$.
Answer: (a)
Given $\vec{r} \times \vec{a} - \vec{r} \times \vec{b} = 0$. Therefore, $\vec{r} \times (\vec{a} - \vec{b}) = 0$. This implies $\vec{r} = \lambda (\vec{a} - \vec{b})$. Thus, $\vec{r} = \lambda (-5\hat{i} - 4\hat{j} + 10\hat{k})$. Also, $\vec{r} \cdot (\hat{i} + 2\hat{j} + \hat{k}) = -3$. Therefore, $\lambda (-5 - 8 + 10) = -3$. Solving gives $\lambda = 1$. Now $\vec{r} = -5\hat{i} - 4\hat{j} + 10\hat{k}$. Then $\vec{r} \cdot (2\hat{i} - 3\hat{j} + \hat{k}) = -10 + 12 + 10 = 12$.
Maths · Straight Lines and Pair of Straight Lines · Single correct
In a triangle PQR, the co-ordinates of the points P and Q are $(-2,4)$ and $(4,-2)$ respectively. If the equation of the perpendicular bisector of PR is $2x - y + 2 = 0$, then the centre of the circumcircle of the $\triangle PQR$ is :
Answer: (b)
Equation of perpendicular bisector of PR is $y = x$. Solving with $2x - y + 2 = 0$ will give $(-2, 2)$.
Maths · Determinants · Single correct
The system of equations $kx + y + z = 1$, $x + ky + z = k$ and $x + y + zk = k^2$ has no solution if $k$ is equal to
Answer: (d)
Given the equations: $$kx + y + z = 1$$ $$x + ky + z = k$$ $$x + y + zk = k^2$$ The determinant $\Delta$ is given by: $$\Delta = \begin{vmatrix} K & 1 & 1 \\ 1 & K & 1 \\ 1 & 1 & K \end{vmatrix} = K \left( K^2 - 1 \right) - 1 \left( K - 1 \right) + 1(1 - K)$$ Simplifying, we have: $$= K^3 - K - K + 1 + 1 - K$$ $$= K^3 - 3K + 2$$ $$= (K - 1)^2 (K + 2)$$ For $K = 1$: $$\Delta = \Delta_1 = \Delta_2 = \Delta_3 = 0$$ But for $K = -2$, at least one out of $\Delta_1, \Delta_2, \Delta_3$ are not zero. Hence for no solution, $K = -2$.
Maths · Inverse Trigonometric Functions · Single correct
If $\cot^{-1}(\alpha) = \cot^{-1} 2 + \cot^{-1} 8 + \cot^{-1} 32 + \ldots$ upto 100 terms, then $\alpha$ is :
Answer: (a)
\[ \begin{aligned} \cot^{-1}(\alpha) &= \cot^{-1}(2)+\cot^{-1}(8)+\cot^{-1}(18)+\cdots \\ &= \sum_{n=1}^{100}\tan^{-1}\!\left(\frac{2}{4n^2-1}\right) \\ &= \sum_{n=1}^{100}\tan^{-1}\!\left(\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}\right) \\ &= \sum_{n=1}^{100}\Bigl[\tan^{-1}(2n+1)-\tan^{-1}(2n-1)\Bigr] \\ &= \tan^{-1}(201)-\tan^{-1}(1) \\ &= \tan^{-1}\!\left(\frac{200}{202}\right) \\ \therefore\quad \cot^{-1}(\alpha) &= \cot^{-1}\!\left(\frac{202}{200}\right) \\ \alpha &= 1.01. \end{aligned} \]
Maths · Three Dimensional Geometry · Single correct
The equation of the plane which contains the y-axis and passes through the point (1,2,3) is :
Answer: (d)
The vector $\vec{n} = \hat{j} \times (\hat{i} + 2\hat{j} + 3\hat{k})$ is calculated as follows: $$\vec{n} = -3\hat{i} + 0\hat{j} + \hat{k}$$ So, $(-3)(x - 1) + 0(y - 2) + (1)(z - 3) = 0$ simplifies to: $$-3x + z = 0$$ Option 4 Alternate: The required plane is given by the determinant: $$\begin{vmatrix} x & y & z \\ 0 & 1 & 0 \\ 1 & 2 & 3 \end{vmatrix} = 0$$ This simplifies to: $$3x - z = 0$$
Maths · Matrices · Single correct
If $\mathbf{A} = \begin{pmatrix} 0 & \sin \alpha \\ \sin \alpha & 0 \end{pmatrix}$ and $\det \left( \mathbf{A}^2 - \frac{1}{2} \mathbf{I} \right) = 0$, then a possible value of $\alpha$ is
Answer: (c)
$A^2=\sin^2\alpha\,I$ So, $\left|A^2-\frac{1}{2}I\right|=\left(\sin^2\alpha-\frac{1}{2}\right)^2$ $=0$ $\Rightarrow \sin\alpha=\pm\frac{1}{\sqrt{2}}$
Maths · Mathematical Reasoning · Single correct
If the Boolean expression $(p \Rightarrow q) \Leftrightarrow (q * (\sim p))$ is a tautology, then the Boolean expression $p * (\sim q)$ is equivalent to
Answer: (a)
Given $\cdot : p \rightarrow q \equiv \sim p \lor q$. So, $\ast \equiv \lor$. Thus, $\mathbf{p}^\ast (\sim q) \equiv p \lor (\sim q)$ which is equivalent to $q \rightarrow p$.
Maths · Probability · Single correct
Two dices are rolled. If both dices have six faces numbered 1, 2, 3, 5, 7 and 11, then the probability that the sum of the numbers on the top faces is less than or equal to 8 is:
Answer: (b)
Given $n(E) = 5 + 4 + 4 + 3 + 1 = 17$. So, $P(E) = \frac{17}{36}$.
Maths · Binomial Theorem · Single correct
If the fourth term in the expansion of $\left( x + x^{\log_2 x} \right)^7$ is 4480, then the value of $x$ where $x \in \mathbb{N}$ is equal to:
Answer: (a)
Given $$^7C_3 x^4 x^{\left(3 \log_2^2 \right)} = 4480$$. Therefore, $$x^{\left(4 + 3 \log_2^2 \right)} = 2^7$$. This implies $$(4 + 3t)t = 7; \ t = \log_2 x$$. Thus, $$t = 1, \ -\frac{7}{3} \Rightarrow x = 2$$.
Maths · Sets · Single correct
In a school, there are three types of games to be played. Some of the students play two types of games, but none play all the three games. Which Venn diagrams can justify the above statement?
Answer: (c)
A $\cap$ B $\cap$ C is visible in all three Venn diagram. Hence, Option (3)
Maths · Inverse Trigonometric Functions · Single correct
The sum of possible values of $x$ for $\tan^{-1}(x+1) + \cot^{-1}\left(\frac{1}{x-1}\right) = \tan^{-1}\left(\frac{8}{31}\right)$ is
Answer: (a)
Given $$\tan^{-1}(x+1) + \cot^{-1}\left(\frac{1}{x-1}\right) = \tan^{-1}\frac{8}{31}$$ Taking tangent both sides: $$\frac{(x+1)+(x-1)}{1-(x^2-1)} = \frac{8}{31}$$ $$\Rightarrow \frac{2x}{2-x^2} = \frac{8}{31}$$ $$\Rightarrow 4x^2 + 31x - 8 = 0$$ $$\Rightarrow x = -8, \frac{1}{4}$$ But, if $x = \frac{1}{4}$ $$\tan^{-1}(x+1) \in \left(0, \frac{\pi}{2}\right)$$ $$\& \cot^{-1}\left(\frac{1}{x-1}\right) \in \left(\frac{\pi}{2}, \pi\right)$$ $$\Rightarrow LHS > \frac{\pi}{2} \& RHS < \frac{\pi}{2}$$ (Not possible) Hence, $x = -8$
Maths · Complex Numbers and Quadratic Equations · Single correct
The area of the triangle with vertices A(z), B(iz) and C(z + iz) is :
Answer: (c)
The area $A$ is given by the formula: $$A = \frac{1}{2} |z| |iz|$$ Simplifying, we have: $$= \frac{|z|^2}{2}$$
Maths · Conic Sections · Single correct
The line $2x - y + 1 = 0$ is a tangent to the circle at the point $(2,5)$ and the centre of the circle lies on $x - 2y = 4$. Then, the radius of the circle is:
Answer: (a)
Given the equation $$\left(\frac{h - \frac{h - 4}{2}}{2 - h}\right)(2) = -1$$ we find that $$h = 8$$. The center is at $$(8, 2)$$. The radius is calculated as $$\sqrt{(8 - 2)^2 + (2 - 5)^2} = 3\sqrt{5}$$.
Maths · Permutations and Combinations · Single correct
Team 'A' consists of 7 boys and $n$ girls and Team 'B' has 4 boys and 6 girls. If a total of 52 single matches can be arranged between these two teams when a boy plays against a boy and a girl plays against a girl, then $n$ is equal to:
Answer: (c)
Total matches between boys of both team $$= {^7C_1} \times {^4C_1} = 28$$ Total matches between girls of both team $$= {^nC_1} {^6C_1} = 6n$$ Now, $28 + 6n = 52$ $$\Rightarrow n = 4$$
Maths · Sequences and Series · Single correct
The value of $4+\cfrac{1}{5+\cfrac{1}{4+\cfrac{1}{5+\cfrac{1}{4+\cdots\infty}}}}$
Answer: (a)
Given the equation: $$y = 4 + \frac{1}{\left(5 + \frac{1}{y}\right)}$$ Subtract 4 from both sides: $$y - 4 = \frac{y}{(5y + 1)}$$ Multiply both sides by $(5y + 1)$: $$5y^2 - 20y - 4 = 0$$ Solve the quadratic equation: $$y = \frac{20 + \sqrt{480}}{10}$$ $$y = \frac{20 - \sqrt{480}}{10} \rightarrow rejected$$ Thus, $$y = 2 + \sqrt{\frac{480}{100}}$$ Correct with Option (A)
Maths · Conic Sections · Single correct
Choose the incorrect statement about the two circles whose equations are given below: $$x^2 + y^2 - 10x - 10y + 41 = 0$$ and $$x^2 + y^2 - 16x - 10y + 80 = 0$$
Answer: (b)
Given $r_1 = 3$, $c_1(5, 5)$ and $r_2 = 3$, $c_2(8, 5)$. The distance $C_1C_2 = 3$, $r_1 = 3$, $r_2 = 3$.
Maths · Applications of Derivatives · Single correct
Which of the following statements is correct for the function $g(\alpha)$ for $\alpha \in \mathbb{R}$ such that $$g(\alpha) = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin^{\alpha} x}{\cos^{\alpha} x + \sin^{\alpha} x} \, dx$$
Answer: (d)
Given $$g(\alpha) = \int_{\pi/6}^{\pi/3} \frac{\sin^\alpha x}{(\sin^\alpha x + \cos^\alpha x)} \cdots (i)$$ $$g(\alpha) = \int_{\pi/6}^{\pi/3} \frac{\cos^\alpha x}{(\sin^\alpha x + \cos^\alpha x)} \cdots (ii)$$ Adding (1) and (2), $$2g(\alpha) = \frac{\pi}{6}$$ Thus, $$g(\alpha) = \frac{\pi}{12}$$ Constant and even function. Due to typing mistake it must be bonus.
Maths · Differential Equations · Single correct
Which of the following is true for $y(x)$ that satisfies the differential equation $$\frac{dy}{dx} = xy - 1 + x - y; \; y(0) = 0$$
Answer: (a)
Given \[ \frac{dy}{dx}=(1+y)(x-1) \] \[ \frac{dy}{y+1}=(x-1)\,dx \] Integrate \[ \ln(y+1)=\frac{x^2}{2}-x+c \] \[ (0,0)\Rightarrow c=0 \] \[ \Rightarrow\quad y=e^{\left(\frac{x^2}{2}-x\right)}-1 \]
Maths · Limits and Derivatives · Single correct
The value of $\lim_{x \to 0^+} \frac{\cos^{-1}(x - [x]^2) \cdot \sin^{-1}(x - [x]^2)}{x - x^3}$, where $[x]$ denotes the greatest integer $\leq x$ is:
Answer: (d)
The limit to evaluate is $$\lim_{x \to 0^+} \frac{\cos^{-1} x}{(1-x^2)} \times \frac{\sin^{-1} x}{x} = \frac{\pi}{2}$$.
Maths · Linear Programming · Numerical
The maximum value of $z$ in the following equation $z = 6xy + y^2$, where $3x + 4y \leq 100$ and $4x + 3y \leq 75$ for $x \geq 0$ and $y \geq 0$ is ____ (Round off to the Nearest Integer)
Answer: 904
Given the equations: $$z = 6xy + y^2 = y(6x + y)$$ Subject to the constraints: $$3x + 4y \leq 100$$ $$4x + 3y \leq 75$$ $$x \leq \left( i \right)$$ The maximum value of $Z$ is given by: $$Z \leq \frac{1}{2} \left( 225y - 7y^2 \right) \leq \frac{(225)^2}{2 \times 4 \times 7}$$ Calculating: $$= \frac{50625}{56}$$ $$\approx 904.0178$$ $$\approx 904.02$$ It will be attained at $y = \frac{225}{14}$.
Maths · Continuity and Differentiability · Numerical
If the function $f(x) = \frac{\cos(\sin x) - \cos x}{x^4}$ is continuous at each point in its domain and $f(0) = \frac{1}{k}$, then $k$ is
Answer: 6
Given $\lim_{x \to 0} \frac{\cos(\sin x) - \cos x}{x^4} = f(0)$. $$\Rightarrow \lim_{x \to 0} \frac{2 \sin \left( \frac{\sin x + x}{2} \right) \sin \left( \frac{\sin x - x}{2} \right)}{x^4} = \frac{1}{K}$$ $$\Rightarrow \lim_{x \to 0} 2 \left( \frac{\sin x + x}{2x} \right) \left( \frac{x - \sin x}{2x^3} \right) = \frac{1}{K}$$ $$\Rightarrow 2 \times \frac{(1+1)}{2} \times \frac{1}{2} \times \frac{1}{6} = \frac{1}{K}$$ $$\Rightarrow K = 6$$
Maths · Continuity and Differentiability · Numerical
If $f(x) = \sin \left( \cos^{-1} \left( \frac{1 - 2^x}{1 + 2^x} \right) \right)$ and its first derivative with respect to $x$ is $-\frac{b}{a} \log_e 2$ when $x = 1$, where $a$ and $b$ are integers, then the minimum value of $|a^2 - b^2|$ is ____
Answer: 481
Given $f(x) = \sin \left( \cos^{-1} \left( \frac{1 - 2^{2x}}{1 + 2^{2x}} \right) \right)$ at $x = 1$; $2^{2x} = 4$. For $\sin \left( \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \right)$. Let $\tan^{-1} x = \theta$; $\theta \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$. Therefore, $\sin (\cos^{-1} \cos 2\theta) = \sin 2\theta$. [ \begin{cases} \text{If } x > 1 \implies \frac{\pi}{2} > \theta > \frac{\pi}{4} \ \therefore \pi > 2\theta > \frac{\pi}{2} \end{cases} ] = $2 \sin \theta \cos \theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}$. Therefore, $f'(1) = \frac{20 \ln 2 - 32 \ln 2}{25} = -\frac{12}{25} \ln 2$. So, $a = 25$, $b = 12 \implies |a^2 - b^2| = 25^2 - 12^2$. $= 625 - 144$. $= 481$.
Maths · Probability (Advanced) · Fill in the blank
Let there be three independent events $E_1$, $E_2$ and $E_3$. The probability that only $E_1$ occurs is $\alpha$, only $E_2$ occurs is $\beta$ and only $E_3$ occurs is $\gamma$. Let $p$ denote the probability that none of the events occurs, satisfying the equations $$ (\alpha-2\beta)p=\alpha\beta $$ and $$ (\beta-3\gamma)p=2\beta\gamma. $$ All the given probabilities are assumed to lie in the interval $(0,1)$. Then, $$ \frac{\mathrm{Probability\ of\ occurrence\ of}\ E_1}{\mathrm{Probability\ of\ occurrence\ of}\ E_3} $$ is equal to $\underline{\hspace{2cm}}$.
Answer: 6
Let $\mathrm{P}(E_1) = P_1$; $\mathrm{P}(E_2) = P_2$; $\mathrm{P}(E_3) = P_3$. $$\mathrm{P}\left(\overline{E_1} \cap \overline{E_2} \cap \overline{E_3}\right) = \alpha = P_1 (1 - P_2) (1 - P_3) \ldots (1)$$ $$\mathrm{P}\left(\overline{E_1} \cap E_2 \cap \overline{E_3}\right) = \beta = (1 - P_1) P_2 (1 - P_3) \ldots (2)$$ $$\mathrm{P}\left(\overline{E_1} \cap \overline{E_2} \cap E_3\right) = \gamma = (1 - P_1) (1 - P_2) P_3 \ldots (3)$$ $$\mathrm{P}\left(\overline{E_1} \cap \overline{E_2} \cap \overline{E_3}\right) = P = (1 - P_1) (1 - P_2) (1 - P_3) \ldots (4)$$ Given that, $(\alpha - 2\beta)P = \alpha \beta$ $$\left(P_1 (1 - P_2) (1 - P_3) - 2 (1 - P_1) P_2 (1 - P_3)\right) P = P_1 P_2$$ $$\Rightarrow (P_1 (1 - P_2) - 2 (1 - P_1) P_2) = P_1 P_2$$ $$\Rightarrow (P_1 - P_1 P_2 - 2 P_2 + 2 P_1 P_2) = P_1 P_2$$ $$\Rightarrow P_1 = 2 P_2 \ldots (1)$$ and similarly, $(\beta - 3 \gamma) P = 2 B \gamma$ $$P_2 = 3 P_3 \ldots (2)$$ So, $P_1 = 6 P_3 \Rightarrow \frac{P_1}{P_3} = 6$
Maths · Vector Algebra · Fill in the blank
If $\vec{a}=\alpha\hat{i}+\beta\hat{j}+3\hat{k}$, $\vec{b}=-\beta\hat{i}-\alpha\hat{j}-\hat{k}$, and $\vec{c}=\hat{i}-2\hat{j}-\hat{k}$, such that $\vec{a}\cdot\vec{b}=1$ and $\vec{b}\cdot\vec{c}=-3$, then $\frac{1}{3}\left((\vec{a}\times\vec{b})\cdot\vec{c}\right)$ is equal to $\underline{\hspace{2cm}}$.
Answer: 2
Given $\vec{a} \cdot \vec{b} = 1 \Rightarrow -\alpha \beta - \alpha \beta - 3 = 1$. This implies $-2 \alpha \beta = 4 \Rightarrow \alpha \beta = -2$. Also, $\vec{b} \cdot \vec{c} = -3 \Rightarrow -\beta + 2 \alpha + 1 = -3$. Solving (1) and (2), $(\alpha, \beta) = (-1, 2)$. $$\frac{1}{3} \left[ \begin{array}{ccc} \vec{a} & \vec{b} & \vec{c} \end{array} \right] = \frac{1}{3} \begin{vmatrix} \alpha & \beta & 3 \\ -\beta & -\alpha & -1 \\ 1 & -2 & -1 \end{vmatrix}$$ $$= \frac{1}{3} \begin{vmatrix} -1 & 2 & 3 \\ -2 & 1 & -1 \\ 1 & -2 & -1 \end{vmatrix}$$ $$= \frac{1}{3} \begin{vmatrix} -2 & 1 & -1 \\ 1 & -2 & -1 \end{vmatrix} = \frac{1}{3} [2(4 - 1)] = 2$$
Maths · Matrices · Numerical
If $A = \begin{bmatrix} 2 & 3 \\ 0 & -1 \end{bmatrix}$, then the value of $\det \left( A^4 \right) + \det \left( A^{10} - (Adj(2\, A))^{10} \right)$ is equal to
Answer: 16
Given $2A adj (2A) = 12AI$. Therefore, $A adj (2A) = -4I \ldots (i)$ Now, $E = \left| A^4 \right| + \left| A^{10} - (adj (2A))^{10} \right|$ $$= (-2)^4 + \frac{\left| A^{20} - A^{10} (adj 2A)^{10} \right|}{\left| A \right|^{10}}$$ $$= 16 + \frac{\left| A^{20} - (A adj (2A))^{10} \right|}{\left| A \right|^{10}}$$ $$= 16 + \frac{\left| A^{20} - 2^{10} I \right|}{2^{10}} (from $$ Now, characteristic roots of $A$ are $2$ and $-1$. So, characteristic roots of $A^{20}$ are $2^{10}$ and $1$. Hence, $\left( A^{20} - 2^{10} I \right) \left( A^{20} - I \right) = 0$ $$\Rightarrow \left| A^{20} - 2^{10} I \right| = 0 (as A^{20} \neq I)$$ Therefore, $E = 16$. Ans.
Maths · Integrals · Numerical
If [.] represents the greatest integer function, then the value of $$\left| \int_{0}^{\frac{\pi}{2}} \left[ x^2 \right] - \cos x \, dx \right|$$ is ____
Answer: 1
Given $$I = \int_0^{\sqrt{\pi/2}} \left( \lfloor x^2 \rfloor + \lfloor -\cos x \rfloor \right) dx$$ This can be split into three integrals: $$= \int_0^1 0 \, dx + \int_1^{\sqrt{\pi/2}} 1 \, dx + \int_0^{\sqrt{\pi/2}} (-1) \, dx$$ Calculating each integral, we have: $$= \sqrt{\frac{\pi}{2}} - 1 - \sqrt{\frac{\pi}{2}} = -1$$ Thus, $$\Rightarrow |I| = 1$$
Maths · Conic Sections · Fill in the blank
The minimum distance between any two points $P_1$ and $P_2$ while considering point $P_1$ on one circle and point $P_2$ on the other circle for the given circles' equations $$x^2 + y^2 - 10x - 10y + 41 = 0$$ $$x^2 + y^2 - 24x - 10y + 160 = 0$$ is .
Answer: 1
Given $C_1(5, 5)$, $r_1 = 3$ and $C_2(12, 5)$, $r_2 = 3$. Now, $C_1C_2 > r_1 + r_2$. Thus, $(P_1P_2)_{\min} = 7 - 6 = 1$.
Maths · Three Dimensional Geometry · Numerical
If the equation of the plane passing through the line of intersection of the planes $$2x - 7y + 4z - 3 = 0, 3x - 5y + 4z + 11 = 0$$ and the point $(-2,1,3)$ is $ax + by + cz - 7 = 0$, then the value of $2a + b + c - 7$ is .
Answer: 4
Required plane is $\mathbf{p}_1 + \lambda \mathbf{p}_2 = (2 + 3\lambda)x - (7 + 5\lambda)y + (4 + 4\lambda)z - 3 + 11\lambda = 0$ which is satisfied by $(-2, 1, 3)$. Hence, $\lambda = \frac{1}{6}$. Thus, plane is $15x - 47y + 28z - 7 = 0$. So, $2a + b + c - 7 = 4$
Maths · Binomial Theorem · Numerical
If $(2021)^{3762}$ is divided by 17, then the remainder is ____.
Answer: 4
Given $$(2023 - 2)^{3762} = 2023k_1 + 2^{3762}$$ $$= 17k_2 + 2^{3762} (as 2023 = 17 \times 17 \times 9)$$ $$= 17k_2 + 4 \times 16^{940}$$ $$= 17k_2 + 4 \times (17 - 1)^{940}$$ $$= 17k_2 + 4 \left(17k_3 + 1\right)$$ $$= 17k + 4 \Rightarrow remainder = 4$$
Physics · System of Particles and Rotational Motion · Single correct
A triangular plate is shown. A force $\vec{F} = 4\hat{i} - 3\hat{j}$ is applied at point P. The torque at point P with respect to point 'O' and 'Q' are:
Answer: (a)
Given $\vec{F} = 4\hat{i} - 3\hat{j}$. $\vec{r}_1 = 5\hat{i} + 5\sqrt{3}\hat{j}$ and $\vec{r}_2 = -5\hat{i} + 5\sqrt{3}\hat{j}$. Torque about 'O' $$\vec{\tau}_O = \vec{r}_1 \times \vec{F} = (-15 - 20\sqrt{3})\hat{k} = (15 + 20\sqrt{3})(-\hat{k})$$ Torque about 'Q' $$\vec{\tau}_Q = \vec{r}_2 \times \vec{F} = (-15 + 20\sqrt{3})\hat{k} = (15 - 20\sqrt{3})(-\hat{k})$$
Physics · Mechanical Properties of Fluids · Single correct
When two soap bubbles of radii $a$ and $b$ $(b > a)$ coalesce, the radius of curvature of common surface is:
Answer: (a)
Excess pressure at common surface is given by $$P_{ex} = 4T \left( \frac{1}{a} - \frac{1}{b} \right) = \frac{4T}{r}$$ Therefore, $$\frac{1}{r} = \frac{1}{a} - \frac{1}{b}$$ $$r = \frac{ab}{b-a}$$
Physics · Kinetic Theory · Single correct
A polyatomic ideal gas has 24 vibrational modes. What is the value of $\gamma$?
Answer: (a)
Since each vibrational mode has 2 degrees of freedom hence total vibrational degrees of freedom = 48 $$f = 3 + 3 + 48 = 54$$ $$\gamma = 1 + \frac{2}{f} = \frac{28}{27} = 1.03$$
Physics · Atoms · Single correct
If an electron is moving in the $n^{th}$ orbit of the hydrogen atom, then its velocity $(v_n)$ for the $n^{th}$ orbit is given as:
Answer: (b)
We know velocity of electron in $n^{th}$ shell of hydrogen atom is given by $$v = \frac{2\pi k Z e^2}{n h}$$ Therefore, $v \propto \frac{1}{n}$
Physics · Dual Nature of Radiation and Matter · Single correct
An electron of mass $m$ and a photon have same energy $E$. The ratio of wavelength of electron to that of photon is: (c being the velocity of light)
Answer: (b)
Given $\($ $\lambda$_1 = $\frac{h}{\sqrt{2mE}}$ $\)$ and $\($ $\lambda$_2 = $\frac{hc}{E}$ $\)$. The ratio $\($ $\frac{\lambda_1}{\lambda_2}$ = $\frac{1}{c}$ $\left$( $\frac{E}{2m}$ $\right$)^{1/2} $\)$.
Physics · Thermal Properties of Matter · Single correct
Two identical metal wires of thermal conductivities $K_1$ and $K_2$ respectively are connected in series. The effective thermal conductivity of the combination is:
Answer: (a)
The effective resistance is given by $$R_{eff} = \frac{l}{K_1 \, A} + \frac{l}{K_2 \, A} = \frac{2l}{K_{eq} \, A}$$ The equivalent thermal conductivity is $$K_{eq} = \frac{2 \, K_1 \, K_2}{K_1 + K_2}$$
Physics · Experimental Physics · Single correct
The vernier scale used for measurement has a positive zero error of 0.2 mm. If while taking a measurement it was noted that '0' on the vernier scale lies between 8.5 cm and 8.6 cm, vernier coincidence is 6, then the correct value of measurement is ___ cm. (least count = 0.01 cm)
Answer: (b)
Positive zero error = 0.2 \, $\mathrm{mm}$ Main scale reading = 8.5 \, $\mathrm{cm}$ Vernier scale reading = 6 $\times$ 0.01 = 0.06 \, $\mathrm{cm}$ Final reading = 8.5 + 0.06 - 0.02 = 8.54 \, $\mathrm{cm}$
Physics · Alternating Current · Single correct
An AC current is given by $I = I_1 \sin \omega t + I_2 \cos \omega t$. A hot wire ammeter will give a reading:
Answer: (b)
Given $I = I_1 \sin \omega t + I_2 \cos \omega t$. Therefore, $I_0 = \sqrt{I_1^2 + I_2^2}$. Thus, $I_{rms} = \frac{I_0}{\sqrt{2}} = \sqrt{\frac{I_1^2 + I_2^2}{2}}$.
Physics · Laws of Motion · Single correct
A modern grand-prix racing car of mass $m$ is travelling on a flat track in a circular arc of radius $R$ with a speed $v$. If the coefficient of static friction between the tyres and the track is $\mu_s$, then the magnitude of negative lift $F_L$ acting downwards on the car is: (Assume forces on the four tyres are identical and $g = acceleration due to gravity$)
Answer: (b)
Given $\mu_s N = \frac{mv^2}{R}$. We have $$N = \frac{mv^2}{\mu_s R} = mg + F_L$$ Therefore, $$F_L = \frac{mv^2}{\mu_s R} - mg$$
Physics · Motion in a Straight Line · Single correct
A car accelerates from rest at a constant rate $\alpha$ for some time after which it decelerates at a constant rate $\beta$ to come to rest. If the total time elapsed is $t$ seconds, the total distance travelled is:
Answer: (c)
Given $v_0 = \alpha t_1$ and $0 = v_0 - \beta t_2 \Rightarrow v_0 = \beta t_2$. $t_1 + t_2 = t$. From the graph, $$v_0 \left( \frac{1}{\alpha} + \frac{1}{\beta} \right) = t$$ $$\Rightarrow v_0 = \frac{\alpha \beta t}{\alpha + \beta}$$ Distance is the area of the $v-t$ graph. $$= \frac{1}{2} \times t \times v_0 = \frac{1}{2} \times t \times \frac{\alpha \beta t}{\alpha + \beta} = \frac{\alpha \beta t^2}{2(\alpha + \beta)}$$
Physics · Moving Charges and Magnetism · Single correct
A solenoid of 1000 turns per metre has a core with relative permeability 500. Insulated windings of the solenoid carry an electric current of 5 A. The magnetic flux density produced by the solenoid is: (permeability of free space = $4\pi \times 10^{-7} \mathrm{H/m}$)
Answer: (a)
Given $B = \mu n I = \mu_0 \mu_{rn} I$. $$B = 4 \pi \times 10^{-7} \times 500 \times 1000 \times 5$$ $$B = \pi Tesla$$
Physics · System of Particles and Rotational Motion · Single correct
A mass $M$ hangs on a massless rod of length $l$ which rotates at a constant angular frequency. The mass $M$ moves with steady speed in a circular path of constant radius. Assume that the system is in steady circular motion with constant angular velocity $\omega$. The angular momentum of $M$ about point $A$ is $L_A$ which lies in the positive $z$ direction and the angular momentum of $M$ about $B$ is $L_B$. The correct statement for this system is :
Answer: (d)
We know, $\vec{L} = m(\vec{r} \times \vec{v})$. Now with respect to A, we always get direction of $\vec{L}$ along the z-axis and also constant magnitude as $mvr$. But with respect to B, we get constant magnitude but continuously changing direction.
Physics · Oscillations · Single correct
For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal?
Answer: (c)
KE = PE $$\frac{1}{2} m \omega^2 \left( A^2 - x^2 \right) = \frac{1}{2} m \omega^2 x^2$$ $$A^2 - x^2 = x^2$$ $$2x^2 = A^2$$ $$x = \pm \frac{A}{\sqrt{2}}$$
Physics · Thermodynamics · Single correct
A Carnot's engine working between 400 K and 800 K has a work output of 1200 J per cycle. The amount of heat energy supplied to the engine from the source in each cycle is :
Answer: (d)
Physics · Ray Optics and Optical Instruments · Single correct
The thickness at the centre of a plano convex lens is 3 mm and the diameter is 6 cm. If the speed of light in the material of the lens is $2 \times 10^8 \, \mathrm{ms}^{-1}$. The focal length of the lens is
Answer: (d)
Given $R^2 = r^2 + (R - t)^2$. Expanding, we have $$R^2 = r^2 + R^2 + t^2 - 2Rt.$$ Neglecting $t^2$, we get $$R = \frac{r^2}{2t}.$$ Therefore, $$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right) = \frac{\mu - 1}{R}.$$ Substituting, $$f = \frac{R}{\mu - 1} = \frac{r^2}{2t(\mu - 1)} = \frac{(3 \times 10^{-2})^2}{2 \times 3 \times 10^{-3} \times \left( \frac{3}{2} - 1 \right)}.$$ Simplifying, $$= \frac{9 \times 10^{-4}}{6 \times 10^{-3} \times 1} \times 2.$$ Therefore, $$f = 0.3 \, \mathrm{m} = 30 \, \mathrm{cm}.$$
Physics · Communication Systems · Single correct
The output of the given combination gates represents:
Answer: (b)
By De Morgan's theorem, we have: $$\overline{A \cdot B} = NAND$$
Physics · Work, Energy and Power · Single correct
A boy is rolling a $0.5\,\mathrm{kg}$ ball on a frictionless floor with a speed of $20\,\mathrm{m\,s^{-1}}$. The ball gets deflected by an obstacle on the way. After deflection, it moves with $5\%$ of its initial kinetic energy. What is the speed of the ball now?
Answer: (b)
Given, $m = 0.5 \, \mathrm{kg}$ and $u = 20 \, \mathrm{m/s}$. Initial kinetic energy $(k_i) = \frac{1}{2} mu^2$ $$= \frac{1}{2} \times 0.5 \times 20 \times 20 = 100 \, \mathrm{J}$$ After deflection it moves with 5$\%$ of $k_i$. $$\therefore \, k_f = \frac{5}{100} \times k_i \Rightarrow \frac{5}{100} \times 100$$ $$\Rightarrow k_f = 5 \, \mathrm{J}$$ Now, let the final speed be $v' \, \mathrm{m/s}$, then: $$k_f = 5 = \frac{1}{2} mv^2$$ $$\Rightarrow v^2 = 20$$ $$\Rightarrow v = \sqrt{20} = 4.47 \, \mathrm{m/s}$$
Physics · Atoms · Single correct
Which level of the single ionized carbon has the same energy as the ground state energy of hydrogen atom?
Answer: (b)
Energy of H-atom is $E = -13.6Z^2/n^2$. For H-atom $Z = 1$ and for ground state, $n = 1$. Therefore, $$E = -13.6 \times \frac{1^2}{1^2} = -13.6 \, eV.$$ Now for carbon atom (single ionised), $Z = 6$. $$E = -13.6 \frac{Z^2}{n^2} = -13.6 (given).$$ Therefore, $$n^2 = 6^2 \Rightarrow n = 6.$$
Physics · Kinetic Theory · Single correct
Two ideal polyatomic gases at temperatures $T_1$ and $T_2$ are mixed so that there is no loss of energy. If $F_1$ and $F_2$, $m_1$ and $m_2$, $n_1$ and $n_2$ be the degrees of freedom, masses, number of molecules of the first and second gas respectively, the temperature of mixture of these two gases is:
Answer: (b)
Let the final temperature of the mixture be T. Since, there is no loss in energy, $$\Delta U = 0$$ $$\Rightarrow \frac{F_1}{2} n_1 R \Delta T + \frac{F_2}{2} n_2 R \Delta T = 0$$ $$\Rightarrow \frac{F_1}{2} n_1 R \left(T_1 - T\right) + \frac{F_2}{2} n_2 R \left(T_2 - T\right) = 0$$ $$\Rightarrow T = \frac{F_1 n_1 R T_1 + F_2 n_2 R T_2}{F_1 n_1 R + F_2 n_2 R} \Rightarrow \frac{F_1 n_1 T_1 + F_2 n_2 T_2}{F_1 n_1 + F_2 n_2}$$
Physics · Current Electricity · Single correct
A current of 10 A exists in a wire of crosssectional area of 5 $\mathrm{mm}^2$ with a drift velocity of $2 \times 10^{-3} \, \mathrm{ms}^{-1}$. The number of free electrons in each cubic meter of the wire is
Answer: (b)
Given $i = 10 \, \mathrm{A}$, $A = 5 \, \mathrm{mm^2} = 5 \times 10^{-6} \, \mathrm{m^2}$ and $v_d = 2 \times 10^{-3} \, \mathrm{m/s}$. We know, $i = neAv_d$. Therefore, $$10 = n \times 1.6 \times 10^{-19} \times 5 \times 10^{-6} \times 2 \times 10^{-3}$$ $$\Rightarrow n = 0.625 \times 10^{28} = 625 \times 10^{25}$$
Physics · Communication Systems · Numerical
For VHF signal broadcasting,_____ $\mathrm{km}^2$ of maximum service area will be covered by an antenna tower of height $30 \, \mathrm{m}$, if the receiving antenna is placed at ground. Let radius of the earth be $6400 \, \mathrm{km}$. (Round off to the Nearest Integer) (Take $\pi$ as $3.14$)
Answer: 1206
Given $d = \sqrt{2Rh}$. $A = \pi d^2$. $A = \pi 2Rh$. $= 3.14 \times 2 \times 6400 \times \frac{30}{1000}$. $A = 1205.76 \, \mathrm{km^2}$. $A = 1206 \, \mathrm{km^2}$.
Physics · System of Particles and Rotational Motion · Numerical
The angular speed of truck wheel is increased from 900 rpm to 2460 rpm in 26 seconds. The number of revolutions by the truck engine during this time is _____ (Assuming the acceleration to be uniform).
Answer: 728
We know, $\theta = \left( \frac{\omega_1 + \omega_2}{2} \right) t$. Let number of revolutions be $N$. Therefore, $2 \pi N = 2 \pi \left( \frac{900 + 2460}{60 \times 2} \right) \times 26$. $N = 728$.
Physics · Current Electricity · Numerical
The equivalent resistance of series combination of two resistors is 's'. When they are connected in parallel, the equivalent resistance is 'p'. If $s = np$, then the minimum value for $n$ is (Round off to the Nearest Integer)
Answer: 4
Given $R_1 + R_2 = s \ldots (1)$ $$\frac{R_1 R_2}{R_1 + R_2} = p \ldots$$ $$R_1 R_2 = sp$$ $$R_1 R_2 = np^2$$ $$R_1 + R_2 = \frac{n R_1 R_2}{(R_1 + R_2)}$$ $$\frac{(R_1 + R_2)^2}{R_1 R_2} = n$$ For minimum value of $n$ $$R_1 = R_2 = R$$ $$\therefore \; n = \frac{(2R)^2}{R^2} = 4$$
Physics · Electrostatic Potential and Capacitance · Numerical
Four identical rectangular plates with length, $l = 2 \, \mathrm{cm}$ and breadth, $b = \frac{3}{2} \, \mathrm{cm}$ are arranged as shown in figure. The equivalent capacitance between A and C is $\frac{x \varepsilon_0}{d}$. The value of $x$ is (Round off to the Nearest Integer)
Answer: 2
The equivalent capacitance is given by: $$C_{eq} = \frac{2C_0}{3} = \frac{2}{3} \frac{\varepsilon_0 A}{d}$$ Calculating further: $$C_{eq} = \frac{2C_0}{3d} \times \left( 2 \times \frac{3}{2} \right) = 2 \left( \because A = 1 b = 2 \times \frac{3}{2} \right)$$
Physics · Gravitation · Numerical
The radius in kilometer to which the present radius of earth ($R = 6400 \, \mathrm{km}$) to be compressed so that the escape velocity is increased 10 times is
Answer: 64
Given $V_e = \sqrt{\frac{2Gm}{R}}$. $10 \, V_e = \sqrt{\frac{2Gm}{R'}}$. Therefore, $10 = \sqrt{\frac{R}{R'}}$. Thus, $R' = \frac{R}{100} = \frac{6400}{100} = 64 \, km$.
Physics · Oscillations · Numerical
Consider two identical springs each of spring constant $k$ and negligible mass compared to the mass $M$ as shown. Fig. 1 shows one of them and Fig. 2 shows their series combination. The ratios of time period of oscillation of the two SHM is $\frac{T_b}{T_a} = \sqrt{x}$, where value of $x$ is (Round off to the Nearest Integer)
Answer: 2
Given \[ T_a = 2\pi \sqrt{\frac{M}{K}} \] and \[ T_b = 2\pi \sqrt{\frac{M}{K/2}}. \] The ratio is \[ \frac{T_b}{T_a} = \frac{2\pi \sqrt{\frac{M}{K/2}}}{2\pi \sqrt{\frac{M}{K}}} = \sqrt{\frac{K}{K/2}} = \sqrt{2} = \sqrt{x}. \] Therefore, \[ x = 2. \]
Physics · System of Particles and Rotational Motion · Single correct
The following bodies, (1) a ring (2) a disc (3) a solid cylinder (4) a solid sphere, of same mass 'm' and radius 'R' are allowed to roll down without slipping simultaneously from the top of the inclined plane. The body which will reach first at the bottom of the inclined plane is [Mark the body as per their respective numbering given in the question]
Answer: (d)
Given $Mg \sin \theta R = \left( mk^2 + mR^2 \right) \alpha$. $$\alpha = \frac{Rg \sin \theta}{k^2 + R^2} \implies a = \frac{g \sin \theta}{1 + \frac{k^2}{R^2}}$$ $$t = \sqrt{\frac{2s}{a}} = \sqrt{\frac{2s}{g \sin \theta} \left( 1 + \frac{k^2}{R^2} \right)}$$ For least time, $k$ should be least and we know $k$ is least for solid sphere.
Physics · Electrostatic Potential and Capacitance · Numerical
A parallel plate capacitor whose capacitance $C$ is $14 \, \mathrm{pF}$ is charged by a battery to a potential difference $V = 12 \, \mathrm{V}$ between its plates. The charging battery is now disconnected and a porcelain plate with $k = 7$ is inserted between the plates, then the plate would oscillate back and forth between the plates with a constant mechanical energy of ______ pJ (Assume no friction)
Answer: 864
Given $U_i = \frac{1}{2} \times 14 \times 12 \times 12 \, \mathrm{pJ}$ ($\because U = \frac{1}{2} CV^2$) \[ = 1008 \, \mathrm{pJ} \] $U_f = \frac{1008}{7} \, \mathrm{pJ} = 144 \, \mathrm{pJ}$ ($\because C_m = kC_0$) Mechanical energy $= \Delta U$ \[ = 1008 - 144 \] \[ = 864 \, \mathrm{pJ} \]
Physics · Laws of Motion · Numerical
Two blocks ( $m = 0.5 \, \mathrm{kg}$ and $M = 4.5 \, \mathrm{kg}$) are arranged on a horizontal frictionless table as shown in figure. The coefficient of static friction between the two blocks is $\frac{3}{7}$. Then the maximum horizontal force that can be applied on the larger block so that the blocks move together is ______ N. (Round off to the Nearest Integer) [Take $g$ as $9.8 \, \mathrm{m/s^2}$ ]
Answer: 21
Given $a_{max} = \mu g = \frac{3}{7} \times 9.8$. $F = (M + m) a_{max} = 5 a_{max}$. $= 21 Newton$
Physics · Electromagnetic Induction · Numerical
If $2.5 \times 10^{-6} \, \mathrm{N}$ average force is exerted by a light wave on a non-reflecting surface of $30 \, \mathrm{cm}^2$ area during 40 minutes of time span, the energy flux of light just before it falls on the surface is _____ $\mathrm{W/cm}^2$ (Round off to the Nearest Integer) (Assume complete absorption and normal incidence conditions are there)
Answer: 25
Given $F = \frac{IA}{C}$. $$I = \frac{FC}{A} = \frac{2.5 \times 10^{-6} \times 3 \times 10^8}{30} = 25 \, \mathrm{W/cm^2}$$
Chemistry · Surface Chemistry · Single correct
With respect to drug-enzyme interaction, identify the wrong statement:
Answer: (c)
Some drugs do not bind to the enzyme's active site. These bind to a different site of the enzyme which is called the allosteric site. This binding of the inhibitor at the allosteric site changes the shape of the active site in such a way that the substrate cannot recognize it. Such an inhibitor is known as a non-competitive inhibitor.
Chemistry · Hydrocarbons · Single correct
Which of the following is an aromatic compound?
Answer: (a)
The compound shown is an aromatic compound.
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
The product "A" in the above reaction is:
Answer: (b)
The reaction involves the formation of a cyclic acetal from a diketone and ethylene glycol in the presence of an acid catalyst. The ethylene glycol reacts with the diketone to form a five-membered cyclic acetal, releasing ethanol as a byproduct.
Chemistry · Chemical Bonding and Molecular Structure · Single correct
A central atom in a molecule has two lone pairs of electrons and forms three single bonds. The shape of this molecule is:
Answer: (c)
The molecule is $\mathrm{sp^3d}$ hybridised and has a T-shaped geometry.
Chemistry · The d-and f-Block Elements · Single correct
Given below are two statements: Statement I : Potassium permanganate on heating at 573 K forms potassium manganate. Statement II : Both potassium permanganate and potassium manganate are tetrahedral and paramagnetic in nature. In the light of the above statements, choose the most appropriate answer from the options given below:
Answer: (a)
Given the reaction: $$2\mathrm{KMnO_4} \xrightarrow{573 \, \mathrm{K}} \mathrm{K_2MnO_4} + \mathrm{MnO_2} + \mathrm{O_2}$$ Potassium permanganate converts to potassium manganate. Statement-I is correct. Statement-II is incorrect.
Chemistry · Chemistry in Everyday Life · Single correct
Which of the following is correct structure of tyrosine?
Answer: (d)
The structure of Tyrosine amino acid is shown in the image.
Chemistry · Haloalkanes and Haloarenes · Single correct
The above reaction requires which of the following reaction conditions?
Answer: (d)
The reaction shown is the Dow process. The conditions for this process are a temperature of $623 \, \mathrm{K}$ and a pressure of $300 \, \mathrm{atm}$.
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
The absolute value of the electron gain enthalpy of halogens satisfies:
Answer: (c)
Order of electron gain enthalpy (Absolute value) is Cl > F > Br > I.
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
Which of the following compound CANNOT act as a Lewis base?
Answer: (b)
Lewis base: Chemical species which has capability to donate electron pair. In $\mathrm{NF_3}$, $\mathrm{SF_4}$, $\mathrm{ClF_3}$ central atom (i.e. N, S, Cl) having lone pair therefore act as lewis base. In $\mathrm{PCl_5}$ central atom (P) does not have lone pair therefore does not act as lewis base.
Chemistry · Environmental Chemistry · Single correct
Reducing smog is a mixture of:
Answer: (b)
Reducing or classical smog is the combination of smoke, fog and $\mathrm{SO_2}$.
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Hoffmann bromomide degradation of benzamide gives product A, which upon heating with $CHCl_3$ and $NaOH$ gives product B. The structures of A and B are :
Answer: (b)
Hoffmann bromamide degradation reaction: $$\mathrm{C_6H_5CONH_2 + Br_2 + 4NaOH \rightarrow C_6H_5NH_2 (A)}$$ $$\mathrm{C_6H_5NH_2 + CHCl_3/KOH \rightarrow C_6H_5NC (B)}$$ Carbylamine reaction:
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Mesityl oxide is a common name of:
Answer: (d)
IUPAC [4-Methylpent-3-en-2-one]
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Which of the following reaction is an example of ammonolysis?
Answer: (d)
The process of cleavage of the $\mathrm{C} - \mathrm{X}$ bond by Ammonia molecule is known as ammonolysis. Ex: $\mathrm{R} - \mathrm{CH}_2 - \mathrm{Cl} + \mathrm{\ddot{N}H}_3 \rightarrow \mathrm{R} - \mathrm{CH}_2 - \mathrm{NH}_2$
Chemistry · Hydrocarbons · Single correct
Answer: (d)
The reaction begins with the addition of HBr to the alkene in the presence of $CCl_4$, forming a carbocation intermediate. A methyl shift occurs to stabilize the carbocation, resulting in a more stable tertiary carbocation. Finally, the bromide ion attacks the carbocation, leading to the formation of the final product.
Chemistry · The Solid State · Single correct
A colloidal system consisting of a gas dispersed in a solid is called a/an:
Answer: (a)
Colloid of gas dispersed in solid is called solid sol.
Chemistry · Hydrogen · Single correct
The INCORRECT statement(s) about heavy water is (are) $(a)$ used as a moderator in nuclear reactor $(b)$ obtained as a by-product in fertilizer industry. $(c)$ used for the study of reaction mechanism $(d)$ has a higher dielectric constant than water Choose the correct answer from the options given below:
Answer: (c)
The dielectric constant of $\mathrm{H_2O}$ is greater than heavy water.
Chemistry · The s-Block Elements · Single correct
The correct order of conductivity of ions in water is:
Answer: (b)
As the size of gaseous ion decreases, it gets more hydrated in water and hence, the size of aqueous ion increases. When this bulky ion moves in solution, it experiences greater resistance and hence lower conductivity. Size of gaseous ion: $\mathrm{Cs^+ > Rb^+ > K^+ > Na^+}$ Size of aqueous ion: $\mathrm{Cs^+ Rb^+ > K^+ > Na^+}$
Chemistry · Structure of Atom · Single correct
What is the spin-only magnetic moment value (BM) of a divalent metal ion with atomic number 25, in it's aqueous solution?
Answer: (a)
Electronic configuration of divalent metal ion having atomic number 25 is $$\mathrm{Mn^{2+}_{(aq)}} \Rightarrow 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^5$$ Total number of unpaired electrons = 5 $\mu$ (Magnetic moment) $= \sqrt{n(n+2)} \mathrm{BM}$ where $n =$ number of unpaired $e^-$ $$\therefore \mu = \sqrt{5(5+2)} = \sqrt{35} \mathrm{BM} = 5.92 \mathrm{BM}$$
Chemistry · Analytical Chemistry · Single correct
Given below are two statements: Statement-I : Retardation factor ($R_f$) can be measured in meter/centimeter. Statement-II : $R_f$ value of a compound remains constant in all solvents. Choose the most appropriate answer from the options given below:
Answer: (c)
$R_f=\text{retardation factor}$ \[ R_f= \frac{\text{Distance travelled by the substance from reference line (cm)}} {\text{Distance travelled by the solvent from reference line (cm)}} \] Note: $R_f$ value of different compounds are different.
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
The point of intersection and sudden increase in the slope, in the diagram given below, respectively, indicates:
Answer: (a)
At intersection point $\Delta G = 0$ and sudden increase in slope is due to melting or boiling point of the metal.
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Numerical
The reaction of white phosphorus on boiling with alkali in inert atmosphere resulted in the formation of product 'A'. The reaction 1 mol of 'A' with excess of $AgNO_3$ in aqueous medium gives _____ mol(s) of Ag. (Round off to the Nearest Integer).
Answer: 4
Chemistry · Equilibrium · Numerical
0.01 moles of a weak acid HA $(K_a = 2.0 \times 10^{-6})$ is dissolved in $1.0 \, \mathrm{L}$ of $0.1 \, \mathrm{M} \mathrm{HCl}$ solution. The degree of dissociation of HA is ______$\times 10^{-5}$ (Round off to the Nearest Integer). [Neglect volume change on adding HA. Assume degree of dissociation $< 1$]
Answer: 2
The reaction is given by: $$\mathrm{HA} \rightleftharpoons \mathrm{H^+} + \mathrm{A^-}$$ Initial concentrations are $0.01 \, \mathrm{M}$ for $\mathrm{HA}$, $0.1 \, \mathrm{M}$ for $\mathrm{H^+}$, and $0 \, \mathrm{M}$ for $\mathrm{A^-}$. At equilibrium, the concentrations are $(0.01 - x)$ for $\mathrm{HA}$ and $(0.1 + x)$ for $\mathrm{H^+}$, both in $\mathrm{M}$. Approximating, we have $\approx 0.01 \, \mathrm{M}$ for $\mathrm{HA}$ and $\approx 0.1 \, \mathrm{M}$ for $\mathrm{H^+}$. Now, the equilibrium constant $K_a$ is given by: $$K_a = \frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]} \Rightarrow 2 \times 10^{-6} = \frac{0.1 \times x}{0.01}$$ Solving for $x$, we get: $$x = 2 \times 10^{-7}$$ Now, the degree of dissociation $\alpha$ is: $$\alpha = \frac{x}{0.01} = \frac{2 \times 10^{-7}}{0.01} = 2 \times 10^{-5}$$
Chemistry · Structure of Atom · Numerical
A certain orbital has $n = 4$ and $m_L = -3$. The number of radial nodes in this orbital is ____ (Round off to the Nearest Integer).
Answer: 0
Given $n = 4$ and $m_\ell = -3$. Hence, $\ell$ value must be $3$. Now, number of radial nodes $= n - \ell - 1$ $$= 4 - 3 - 1 = 0$$
Chemistry · Some Basic Concepts of Chemistry · Numerical
In the above reaction, 3.9 g of benzene on nitration gives 4.92 g of nitrobenzene. The percentage yield of nitrobenzene in the above reaction is ____ %. (Round off to the Nearest Integer). Given atomic mass : C : 12.0u, H : 1.0u O : 16.0u, N : 14.0u
Answer: 80
1 mole of benzene weighs 78 $\mathrm{gm}$ and 1 mole of nitrobenzene weighs 123 $\mathrm{gm}$. Given 3.9 $\mathrm{gm}$ of benzene, the theoretical yield of nitrobenzene is calculated as follows: $$\frac{123}{78} \times 3.9 = 6.15 \, \mathrm{gm}$$ But the actual amount of nitrobenzene formed is 4.92 $\mathrm{gm}$. Hence, the percentage yield is: $$Percentage yield = \frac{4.92}{6.15} \times 100 = 80\%$$
Chemistry · Some Basic Concepts of Chemistry · Numerical
The mole fraction of a solute in a 100 molal aqueous solution _____ $\times$ $10^{-2}$ (Round off to the Nearest Integer).
Answer: 64
100 molal aqueous solution means there is 100 mole solute in 1 kg = 1000 $\mathrm{gm}$ water. Now, mole-fraction of = $\frac{n_{solute}}{n_{solute} + n_{solvent}}$ $$= \frac{100}{100 + \frac{1000}{18}} = \frac{1800}{2800} = 0.6428$$ $$= 64.28 \times 10^{-2}$$
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
For a certain first order reaction 32$\%$ of the reactant is left after 570 $\mathrm{s}$. The rate constant of this reaction is ____ $\times 10^{-3} \, \mathrm{s}^{-1}$. (Round off to the Nearest Integer). [ Given : $\log_{10} 2 = 0.301$, $\ln 10 = 2.303$ ]
Answer: 2
For first order reaction, $$K = \frac{2.303}{t} \cdot \log \frac{[A_0]}{[A_t]} = \frac{2.303}{570 \, \mathrm{sec}} \cdot \log \left( \frac{100}{32} \right)$$ $$= 1.999 \times 10^{-3} \, \mathrm{sec}^{-1} \approx 2 \times 10^{-3} \, \mathrm{sec}^{-1}$$
Chemistry · Equilibrium · Numerical
The standard enthalpies of formation of Al$_2$O$_3$ and CaO are $-1675 \, \mathrm{kJ \, mol^{-1}}$ and $-635 \, \mathrm{kJ \, mol^{-1}}$ respectively. For the reaction $3\mathrm{CaO} + 2\mathrm{Al} \rightarrow 3\mathrm{Ca} + \mathrm{Al}_2\mathrm{O}_3$ the standard reaction enthalpy $\Delta_r H^0 = \, \mathrm{kJ}$. (Round off to the Nearest Integer).
Answer: 230
Given reaction: $$3\mathrm{CaO} + \mathrm{Al} \rightarrow \mathrm{Al_2O_3} + 3\mathrm{Ca}$$ Now, $\Delta H^\circ = \Sigma \Delta_r H^\circ_{Products} - \Sigma \Delta_r H^\circ$ $$= [1 \times (-1675) + 3 \times 0] - [3 \times (-635) + 2 \times 0]$$ $$= +230 \, \mathrm{kJ \, mol^{-1}}$$
Chemistry · Redox Reactions · Numerical
$15\,\mathrm{mL}$ of aqueous solution of $\mathrm{Fe^{2+}}$ in acidic medium completely reacted with $20\,\mathrm{mL}$ of $0.03\,\mathrm{M}$ aqueous $\mathrm{Cr_2O_7^{2-}}$. The molarity of the $\mathrm{Fe^{2+}}$ solution is \_\_\_\_ $\times 10^{-2}\,\mathrm{M}$ (Round off to the Nearest Integer).
Answer: 24
Given $n_{eq} \mathrm{Fe^{2+}} = n_{eq} \mathrm{Cr_2O_7^{2-}}$. Or, $$\left( \frac{15 \times M_{\mathrm{Fe^{2+}}}}{1000} \right) \times 1 = \left( \frac{20 \times 0.03}{1000} \right) \times 6$$ Therefore, $M_{\mathrm{Fe^{2+}}} = 0.24 \mathrm{M} = 24 \times 10^{-2} \mathrm{M}$.
Chemistry · Solutions · Numerical
The oxygen dissolved in water exerts a partial pressure of 20 $\mathrm{kPa}$ in the vapour above water. The molar solubility of oxygen in water is ____ $\times$ $10^{-5}$ $\mathrm{mol \, dm^{-3}}$ (Round off to the Nearest Integer). [Given : Henry's law constant = $K_H$ = 8.0 $\times$ $10^{4}$ $\mathrm{kPa}$ for $\mathrm{O_2}$ Density of water with dissolved oxygen = 1.0 $\mathrm{kg \, dm^{-3}}$ ]
Answer: 25
Given $P_{(g)} = \left[K_{H}\right] \times 20 \times 10^{3} = \left[8 \times 10^{4}\right]$. Solubility $= \frac{20}{8 \times 10^{4}} = 2.5 \times 10^{-4} = 25 \times 10^{-5}$.
Chemistry · States of Matter · Numerical
The pressure exerted by a non-reactive gaseous mixture of $6.4\,\mathrm{g}$ of methane and $8.8\,\mathrm{g}$ of carbon dioxide in a $10\,\mathrm{L}$ vessel at $27^\circ\mathrm{C}$ is _____ $\mathrm{kPa}$. (Round off to the Nearest Integer) [Assume gases are ideal. $R = 8.314\,\mathrm{J\,mol}^{-1}\,\mathrm{K}^{-1}$ Atomic masses: $\mathrm{C}$: $12.0\,\mathrm{u}$, $\mathrm{H}$: $1.0\,\mathrm{u}$, $\mathrm{O}$: $16.0\,\mathrm{u}$]
Answer: 150
Total moles of gases, $n = n_{\mathrm{CH_4}} + n_{\mathrm{CO_2}}$ $= \frac{6.4}{16} + \frac{8.8}{44} = 0.6$ $P = \frac{nRT}{V} = \frac{0.6 \times 8.314 \times 300}{10 \times 10^{-3}}$ Now, $= 1.49652 \times 10^5\,\mathrm{Pa}$ $= 149.652\,\mathrm{kPa}$ $\approx 150\,\mathrm{kPa}$