JEE Main 7 April 2025 Shift 1 question paper with solutions

The limit to evaluate is given by $$\lim_{x \to 0} \left( \frac{\tan(5x^{1/3})}{5x^{1/3}} \right) \cdot \left( \frac{(3\sqrt{x})^2}{(\tan^{-1} 3\sqrt{x})^2} \right) \cdot \left( \frac{\ell\,(1 + 3x^2)}{3x^2} \right) \cdot \left( \frac{5x^{4/3}}{e^{5x^4} - 1} \right) \times \frac{5x^{1/3} \cdot 3x^2}{5x^{4/3} \cdot 9x}$$ The result of this limit is $$= \frac{1}{3}$$

For line $L_1$: $\($ $\frac{x-1}{2}$ = $\frac{y-2}{3}$ = $\frac{z-3}{4}$ $\)$. For line $L_2$: $\($ $\frac{x}{1}$ = $\frac{y}{\alpha}$ = $\frac{z-5}{1}$ $\)$. The vector $\($ $\vec{x}$ = $\begin{vmatrix}$ $\hat{i}$ & $\hat{j}$ & $\hat{k}$ $\\$ 2 & 3 & 4 $\\$ 1 & $\alpha$ & 1 $\end{vmatrix}$ = $\hat{i}$(3 - 4$\alpha$) - $\hat{j}$(-2) + $\hat{k}$(2$\alpha$ - 3) $\)$. The shortest distance (S.D.) is given by: $$ S.D. = \frac{\vec{BA} \cdot \vec{n}}{\| \vec{n} \|} = \frac{(\hat{i} + 2\hat{j} - 2\hat{k}) \cdot \vec{n}}{\| \vec{n} \|} $$ This implies: $$ 6(13 - 8\alpha)^2 = 25 \left( (4\alpha - 3)^2 + (2\alpha - 3)^2 + 16 \right) $$ Simplifying gives: $$ 6 \left( 64\alpha^2 - 280\alpha + 169 \right) = 25 \left( 20\alpha^2 - 36\alpha + 34 \right) $$ Further simplification leads to: $$ 116\alpha^2 + 348\alpha - 164 = 0 $$ Solving for $\($ $\alpha$ $\)$: $$ \alpha_1 + \alpha_2 = \frac{-348}{116} = -3 $$

Given $f(x) = x^2 + ax^2 + b \ln |x| + 1$, $x \neq 0$. Differentiating, we have $$f'(x) = 3x^2 + 2ax + \frac{b}{x}$$ Evaluating at $x = -1$, $$f'(-1) = 3 - 2a - b = 0$$ Evaluating at $x = -2$, $$f'(-2) = 12 + 4a - \frac{b}{2} = 0$$ Solving these equations, we find $$a = -\frac{9}{2}, b = 12$$ Thus, $$f'(x) = 3x^2 - 9x + \frac{12}{x} = \frac{3(x+1)(x+2)^2}{x}$$ Maximum at $n = -1$. Substituting back, $$f(x) = x^2 - \frac{9}{2}x^2 + 12 \ln |x| + 1$$ At $x = -1$, $$f(-1) = -1 - \frac{9}{2} + 1 = -\frac{9}{2}$$ Maximum value $M = -4.5$. Minimum value at $x = -2$. At $x = -2$, $$f(-2) = -8 - 18 + 12 \ln 2 + 1$$ Thus, $$m = -25 + 12 \ln 2 = -16.6$$ The absolute value of the sum is $$|M + m| = 21.1$$

Let $N = ((64)^{64})^{64}$. $N = (64)^{64^2}$. $N = (1 + 63)^{64^2}$, let $64^2 = n$. Expanding by binomial $$N = (1 + 63)^n = 1 + nC_1 63 + nC_2 (63)^2 + \ldots$$ $$= 1 + 63 \lambda = 1 + 7(9 \lambda)$$ Remainder when divided by 7 is 1

Equation of parabola Given $$(x+2)^2 + (y-1)^2 = \left(\frac{2x+y+2}{\sqrt{5}}\right)^2$$ We have: $$5\left[(x+2)^2 + (y-1)^2\right] = (2x+y+2)^2$$ Put $x = -2$, then: $$5(y-1)^2 = (y-2)^2$$ Expanding both sides: $$5(y^2 - 2y + 1) = y^2 - 4y + 4$$ Simplifying gives: $$\Rightarrow 4y^2 - 6y + 1 = 0 \Rightarrow y_1 + y_2 = \frac{3}{2}$$

Given the equation $x \left( x^2 + e^x \right) \frac{dy}{dx} + \left( e^x (x-2)y - x^3 \right) dx = 0$. Rewriting, we have $x \left( x^2 + e^x \right) \frac{dy}{dx} + e^x (x-2)y = x^3$. This simplifies to $$\frac{dy}{dx} + \frac{e^x (x-2)}{x(x^2 + e^x)} y = \frac{x^2}{x^2 + e^x}.$$ The integrating factor (I.F.) is $e^{\int \frac{e^x (x-2)}{x(x^2 + e^x)} dx} = e^{\int \frac{e^x}{x^2 + e^x} dx}$. Let $1 + \frac{e^x}{x^2} = t \Rightarrow x^2 e^x - e^x 2x dx = x^4 dx = dt$. Thus, $$I.F. = e^{\ln \left( 1 + \frac{e^x}{x^2} \right)} = 1 + \frac{e^x}{x^2}.$$ Now, $$y \left( 1 + \frac{e^x}{x^2} \right) = \int \frac{x^2}{x^2 + e^x} \cdot \frac{x^2 + e^x}{x^2} dx + C.$$ Therefore, $$y \left( 1 + \frac{e^x}{x^2} \right) = x + C.$$ Passing through $(1,0)$ gives $C = -1$. Thus, $$y = \frac{x-1}{1 + \frac{e^x}{x^2}}.$$ Evaluating $y(2)$, we have $$y(2) = \frac{1}{1 + \frac{e^2}{4}} = \frac{4}{4 + e^2}.$$

7 Batsmen and 6 Bowlers. To select 10 players including at least 4 Batsmen and 4 Bowlers. Captain and vice-captain already selected. Number of ways is $$\binom{6}{5} \times \binom{5}{3} + \binom{6}{4} \times \binom{5}{4} + \binom{6}{3} \times \binom{5}{5}$$ $$= 6 \times 10 + 15 \times 5 + 20 \times 1$$ $$= 60 + 75 + 20 = 155$$

Given $$x = 3 \left( \frac{\tan \theta + \sqrt{3}}{1 - \sqrt{3} \tan \theta} \right)$$ $$x - \sqrt{3} \tan \theta = 3 \tan \theta + 3 \sqrt{3}$$ $$\tan \theta = \frac{x - 3 \sqrt{3}}{3 + \sqrt{3} x} \cdots (1)$$ $$2 \left( \frac{\tan \theta + \frac{1}{\sqrt{3}}}{1 - \frac{\tan \theta}{\sqrt{3}}} \right) = y$$ $$2(\sqrt{3} \tan \theta + 1) = y(\sqrt{3} - \tan \theta) \cdots$$ Using (1) and (2) $$2 \left( \frac{x - 3 \sqrt{3}}{\sqrt{3} + x} + 1 \right) = y \left( \sqrt{3} - \frac{x - 3 \sqrt{3}}{\sqrt{3} + x} \right)$$ $$2 \sqrt{3} (x - 3 \sqrt{3} + x + \sqrt{3}) = y(3(\sqrt{3} + x) - x + 3 \sqrt{3})$$ $$4 \sqrt{3} x - 12 = y(2x + 6 \sqrt{3})$$ $$xy - 2 \sqrt{3} x + 3 \sqrt{3} y - 6 = 0$$ $$\Rightarrow \alpha = -2 \sqrt{3}, \beta = 3 \sqrt{3}, \gamma = -6$$ $$\alpha^2 + \beta^2 + \gamma^2 = 12 + 27 + 36 = 75$$

Let $C_1$ and $C_2$ have centres $A(-3, -3)$ and $B(1, 3)$. $AB = \sqrt{16 + 36} = 2\sqrt{13}$ $r_1 = 3$ and $r_2 = 2\sqrt{13} - 3$ $P(\alpha, \beta), \alpha = \frac{r_1(1) + r_2(-3)}{r_1 + r_2}, \beta = \frac{r_1(3) + r_2(-3)}{r_1 + r_2}$ $\alpha = \frac{3 - 3(2\sqrt{13} - 3)}{2\sqrt{13}}, \beta = \frac{18 - 6\sqrt{13}}{2\sqrt{13}}$ $\left( \beta - \alpha \right)^2 = \left( \frac{6}{2\sqrt{13}} \right)^2$ $\left( \beta - \alpha \right)^2 = \left( \frac{6}{2\sqrt{13}} \right)^2, \; m + n = 22$

Given $$I = \int_0^\pi \frac{(x+3) \sin x}{1 + 3 \cos^2 x} \, dx$$ We have $$I = \int_0^\pi \frac{(\pi - x + 3) \sin x}{(1 + 3 \cos^2 x)} \, dx$$ Thus, $$2I = \int_0^\pi \frac{(\pi + 6) \sin x \cdot dx}{(1 + 3 \cos^2 x)} = 2 \int_0^{\pi/2} \frac{(\pi + 6) \sin x}{(1 + 3 \cos^2 x)}$$ Therefore, $$I = \int_0^{\pi/2} \frac{(\pi + 6) \sin x \cdot dx}{(1 + 3 \cos^2 x)} = \frac{\pi}{3\sqrt{3}} (\pi + 6)$$ Let $\sqrt{3} \cos x = t$ and $\sqrt{3} \sin x = dt$.

Given $S_1: |z| = 1$, $\frac{z-i}{z+i} = \frac{\overline{z}+i}{\overline{z}-i}$. This implies $$(z-i)(\overline{z}-i) = (z+i)(\overline{z}+i)$$ $$|z|^2 - i(z + \overline{z}) - 1 = |z|^2 + i(z + \overline{z}) - 1$$ $$i(z + \overline{z}) = 0$$ $$z + \overline{z} = 2 \cos \theta = 0 \Rightarrow \cos \theta = 0$$ $$z = 0 + 0i, \ |z| \neq 1$$ For $S_1: \frac{z-1}{z+1} + \frac{\overline{z}-1}{\overline{z}+1} = 0$, $$(z-1)(\overline{z}+1) + (z+1)(\overline{z}-1) = 0$$ This implies $$|z|^2 + (z - \overline{z}) - 1 + |z|^2 + (\overline{z} - z) - 1 = 0$$ $$|z|^2 = 1$$

Actual means $= \mu = \frac{100(40) - 50 + 40}{100}$. $$\mu = 40 - \frac{1}{10} = 39.9$$ Incorrect variance $$(5.1)^2 = \frac{\sum x_i^2}{100} - \left(\bar{x}\right)^2$$ $$\sum x_i^2 = 100 \times (40^2) + 100(5.1)^2$$ $$\sum x_i^2 = 16 \times 10^4 + (5.1)^2 \times 100 = 162601$$ $$\sigma^2 = \frac{\sum x_i^2 - 50^2 + 40^2}{100} - (\mu)^2$$ $$\sigma^2 = 1617.01 - (39.9)^2 = 25$$ $$\sigma = 5$$ $$10(\mu + \sigma) = 10(39.9 + 5)$$ $$= 10 \times 44.9 = 449$$

Given $x_1, x_2, x_3, x_4 \rightarrow G.P.$ Let $a, ar, ar^2, ar^3 \rightarrow G.P.$ Now $a - 2, ar - 7, a^2 - 9, ar^3 - 5 \rightarrow A.P.$ $2(ar - 7) = a - 2 + ar^2 - 9 \ldots (i)$ $2(a^2 - 9) = ar - 7 + ar^3 - 5 \ldots (ii)$ Solving $r = 2, a = -3$ Therefore, Product $= x_1, x_2, x_3, x_4 = a^4 r^6 = 81 \times 64$ Solving $r = 2, a = -3$ Therefore, Product $= x_1, x_2, x_3, x_4 = a^4 r^6 = 81 \times 64$

Using location of roots: (i) $D \geq 0$ (ii) $\frac{-b}{2a} 0$ $$(p + 2)^2 - 4(2p + 9) \geq 0$$ $$(p + 4)(p - 8) \geq 0 p + 2 0$$ Intersection $p \in \left( -\frac{9}{2}, -4 \right]$ Therefore, $\beta - 2\alpha = -4 + 9 = 5$

Given $|adj(adj(adj A))| = 81$. Therefore, $|adj A|^4 = 81$. This implies $|adj A| = 3$. Hence, $|A|^2 = 3$ and $|A| = \sqrt{3}$. Now, $\($ $\left$( |A|^4 $\right$)^{$\frac{(n-1)^2}{2}$} = |A|^{n^2 - 5n - 4} $\)$. This gives $2(n-1)^2 = 3n^2 - 5n - 4$. Simplifying, $2n^2 - 4n + 2 = 3n^2 - 5n - 4$. Thus, $n^2 - n - 6 = 0$. Factoring, $(n-3)(n+2) = 0$. Therefore, $n = 3, -2$. The sum $\sum_{n \in S} |A^{n^2 + n}| = |A^2| + |A^{12}| = 3 + 36 = 3 + 729 = 732$.

Area is given by $$\int_{-6}^{4} \left\{ \left( 4 - \frac{x^2}{4} \right) - \left( \frac{x - 4}{2} \right) \right\} \, dx$$ which simplifies to $$\int_{-6}^{4} \left\{ -\frac{x^2}{4} - \frac{x - 6}{2} \right\} \, dx$$ Calculating, we find $$\alpha = -\frac{x^3}{12} - 4 + \frac{x^2}{2} + 6x \bigg|_{-6}^{4} = \frac{125}{3}$$ Therefore, $$6\alpha = 250$$

Given the determinant equation: $$\begin{vmatrix} 2 & 3 & 5 \\ 7 & 3 & -2 \\ 12 & 3 & -(\lambda + 4) \end{vmatrix} = 0$$ Expanding the determinant, we have: $$12(-21) - 3(-39) - (\lambda + 4)(-15) = 0$$ Simplifying, we get: $$-252 + 117 + 15(1 + 4) = 0$$ This simplifies to: $$15\lambda + 177 - 252 = 0$$ Solving for $\lambda$, we find: $$15\lambda - 75 = 0 \implies \lambda = 5$$ Next, consider the determinant: $$\begin{vmatrix} 9 & 3 & 5 \\ 8 & 3 & -2 \\ 16 - \mu & 3 & -9 \end{vmatrix} = 0 \implies \begin{vmatrix} 1 & 0 & 7 \\ \mu - 8 & 0 & 7 \\ 16 - \mu & 3 & -9 \end{vmatrix} = 0$$ Expanding, we have: $$7 - 7(\mu - 8) = 0 \implies 1 - (\mu - 8) = 0 \implies \mu = 9$$ Thus, the centre of the circle is $(5, 9)$. The radius is the length of the perpendicular from the centre $(5, 9)$: $$\begin{vmatrix} 20 - 27 \\ 5 \end{vmatrix} = \frac{7}{5}$$

Dr's of AC implies $2\lambda, 3\lambda - 2, 4\lambda$. Dr's of BC implies $\mu + 2, 2\mu + 3, \mu - 1$. $$\frac{\mu + 2}{2\lambda} = \frac{2\mu + 3}{3\lambda - 2} = \frac{\mu - 1}{4\lambda}$$ implies $2(\mu + 2) = \mu - 1$ implies $\mu = -5$. Therefore, Dr's of BC implies $3, 7, 6$. The equation of L implies $$\frac{x - 1}{3} = \frac{y - 1}{7} = \frac{z - 1}{6}$$ $(7, 15, 13)$ satisfies.

Given $\vec{c} = 3 \vec{a} + 6 \vec{b} + 9 (\vec{a} \times \vec{b})$. Since $\sin^{-1}\left(\frac{\sqrt{65}}{9}\right) \Rightarrow \sin \theta = \frac{\sqrt{65}}{9}$, it follows that $\cos \theta = \frac{4}{9}$. Calculate $\vec{c} \cdot \vec{a} = 3 |\vec{a}|^2 + 6 \vec{a} \cdot \vec{b} = 3 + \frac{6 \cdot 4}{9} = \frac{51}{9}$. Next, $\vec{c} \cdot \vec{a} = 3 \vec{a} \cdot \vec{b} + 6 |\vec{b}|^2 = \frac{3 \cdot 4}{9} + 6 = \frac{22}{3}$. Therefore, $9 (\vec{c} \cdot \vec{a}) - 3 (\vec{c} \cdot \vec{b}) = 51 - 22 = 29$.

Equation of Altitude AP: $x = 1$. Equation of Altitude BP: $y - 0 = 1(x + 2)$. Therefore, $x = 1$ and $x - y + 2 = 0$. Point $P(1, 3)$. Area of $\triangle PBC = \frac{1}{2} \times 4 \times 3 = 6$.

Check for $\left[ \frac{x^2}{2} \right]$ and $[\sqrt{x}]$ becomes integers. $\{0, 1, \sqrt{2}, 2, \sqrt{6}, \sqrt{8}, \sqrt{10}, \sqrt{12}, \sqrt{14}, 4\}$ Continuous at $0^+$, continuous at $4^-$ $$\left[ \frac{x^2}{2} \right] = [\sqrt{x}],$$ occurs at $x = \sqrt{2}$ Therefore, not continuous.

Given $A = \{1, 2, 3\}$. The elements $(1, 1), (2, 2), (3, 3), (1, 2) \in R$. Remaining elements are $(2, 1), (2, 3), (1, 3), (3, 1), (3, 2)$. (1) If the relation contains exactly 4 elements, there is 1 way. (2) If the relation contains exactly 5 elements, it can be $(1, 3), (3, 2)$, which gives 2 ways. (3) If the relation contains exactly 6 elements, it can be $((2, 3), (1, 3)), ((1, 3), (3, 2)), ((3, 1), (3, 2))$, which gives 3 ways. Therefore, the total is 6 ways.

Given $f_1 \equiv (-ae, 0) \equiv P(-3, 0)$ and $ae = 3$. We have $\tan 45^\circ = \frac{b^2/a}{2ae}$. Therefore, $2ae = \frac{b^2}{a}$. This implies $b^2 = 6a$. Also, $a^2 e^2 = a^2 + b^2$. Thus, $9 = a^2 + 6a$. Solving $a^2 + 6a - 9 = 0$, we get $a = -3 \pm 3\sqrt{2} = -3(1 \pm \sqrt{2})$. Therefore, $a^2 b^2 = a^2 \cdot 6a = 6a^3 = 6(135\sqrt{2} - 189)$. Let $\alpha = 810$ and $\beta = 1134$. Thus, $\alpha + \beta = 1944$.

Given $\($ $\begin{vmatrix}$ a & d $\\$ b & c $\end{vmatrix}$ = ad - bc $\Rightarrow$ ad = bc $\)$ Case-I Exactly 1 number is used $\($ $\Rightarrow$ $\)$ All singular $\($ = $\binom{4}{1}$ $\)$ Case-II Exactly 2 numbers are used $\($ $\Rightarrow$ $\binom{4}{2}$ $\times$ 2 $\times$ 2 $\)$ Case-III Exactly 3 numbers are used None will be singular Case-IV Exactly 4 numbers are used $\($ ad = bc $\)$ $\($ $\Rightarrow$ 2 $\times$ 9 = 3 $\times$ 6 $\)$ $\($ $\begin{vmatrix}$ 9 & - $\\$ - & 2 $\end{vmatrix}$ $\Rightarrow$ $\binom{4}{1}$ $\times$ 21 $\)$ Total $\($ = 36 $\)$

Given the set $A = \{1, 2, 3, 4, 5, \ldots, n\}$. The number of subsets having $r$ elements such that no two are consecutive is $= n-r+1 C_r$. For $n = 5$, the number of ways is $= {6 \choose r}$. Subsets having no element $= 1$. Subsets having exactly 1 element $= {5 \choose 1} = 5$. Subsets having exactly 2 elements $= {4 \choose 2} = 6$. Subsets having exactly 3 elements $= {3 \choose 3} = 1$. Therefore, $5 + 6 + 1 + 1 = 13$.

The given equation can be written as $$x = \frac{a}{2} \cos [1.5 + 50.5] t + \frac{a}{2} \cos [50.5 - 1.5]$$ $$x = \frac{a}{2} \cos [52t] + \frac{a}{2} \cos [49t]$$ Here, $2 \pi f_1$ and $2 \pi f_2 = 49$ $$f_1 = \frac{52}{2 \pi}, \; f_2 = \frac{49}{2 \pi}$$ Therefore, $f_{Beat} = f_1 - f_2 = \frac{3}{2 \pi} \; Hz$ Thus, $T_{Beat} = \frac{1}{f_{Beat}} = \frac{2 \pi}{3} \; sec$ $$= 2.09 \; sec \approx 2 \; sec$$

Given $$\mathbf{E} = \sqrt{(E_0)^2 + (E_0)^2 + 2\,(E_0)(E_0)\cos\frac{\pi}{3}}$$ We have: $$\mathbf{E} = \sqrt{2E_0^2 + E_0^2} = \sqrt{3}E_0 = 1.73E_0$$

As a balanced Wheatstone bridge is formed, the equivalent resistance between A and B can be written as $$\frac{1}{R_{AB}} = \frac{1}{2r} + \frac{1}{2r} + \frac{1}{r} = \frac{2}{r}$$ Therefore, $$R_{AB} = \frac{R}{12}$$

As $\vec{v}$ is $\perp$ to $\vec{B}$, so charge particle will move in circular path, whose radius is given by $$R = \frac{mv}{qB}$$ Starting point $\rightarrow$ A Ending point $\rightarrow$ C Therefore, Net displacement $=$ AC $$AC = CD - AD$$ $$AC = \frac{2mv}{qB_1} - \frac{2mv}{qB_2}$$ $$AC = \frac{2mv}{qB_1} \left[ 1 - \frac{B_1}{B_2} \right]$$

We know that the formula for displacement current is given by $$i_d = \varepsilon_0 \frac{d\phi_\varepsilon}{dt}$$

The center of mass (COM) of the system is calculated as follows. For the x-coordinate of the COM: $$x_{com} = \frac{2m(10) + 3m(0)}{2m + 3m} = 4 \, cm$$ For the y-coordinate of the COM: $$y_{com} = \frac{5m}{2m + 3m} = 9 \, cm$$ Therefore, the position vector of the COM is: $$\vec{r}_{com} = 4\hat{i} + 9\hat{j}$$

The percentage change in $B$ is given by $$\frac{B_{new} - B_{old}}{B_{old}} \times 100\%$$ Starting with $$\frac{\mu ni - \mu_0 ni}{\mu_0 ni} \times 100\% = \frac{(\mu - \mu_0)}{\mu_0} \times 100\%$$ This simplifies to $$\frac{(\mu_0 \mu_r - \mu_0)}{\mu_0} \times 100\%$$ Which further simplifies to $$(\mu_r - 1) \times 100\%$$ This is equal to $$\chi_n \times 100\%$$ Finally, we have $$= 1.2 \times 10^{-3}\%$$

As we know, $$\frac{1}{f} = \left[ \frac{\mu_L}{\mu_m} - 1 \right] \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$$ For air $\mu_m = 1$ $$\frac{1}{12} = [1.6 - 1] \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$$ $$\frac{1}{12} = \frac{6}{10} \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$$ $$\left[ \frac{1}{R_1} - \frac{1}{R_2} \right] = \frac{10}{72}$$ For water $$\frac{1}{f} = \left[ \frac{1.6}{1.28} - 1 \right] \left[ \frac{10}{72} \right] = \frac{32}{128} \times \frac{10}{72}$$ $$\frac{1}{f} = \frac{1}{4} \times \frac{10}{72}$$ $$f = 28.8 \, \mathrm{cm}$$ $$f = 288 \, \mathrm{mm}$$

Given $i = 5\sqrt{2} + 10 \cos \left(650 \pi t + \frac{\pi}{6}\right)$. $i^2 = 50 + 100 \cos^2 \left(650 \pi t + \frac{\pi}{6}\right) + (2)(5\sqrt{2})(10) \cos \left(650 \pi t + \frac{\pi}{6}\right)$. $\langle i^2 \rangle = 50 + \frac{100}{2} + 0$. $\langle i^2 \rangle = 100$. $\langle i \rangle = 10 Amp$.

Given $f_1 = 30 \, \mathrm{cm}$, $f_2 = 10 \, \mathrm{cm}$. $$\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}, \; d = distance between lens$$ $$\frac{1}{f_{eq}} = \frac{1}{0.3} + \frac{1}{0.1} - \frac{0.1}{(0.3)(0.1)}$$ $$\frac{1}{f_{eq}} = 1 + 0.1$$ $$\frac{1}{f_{eq}} = 1$$ Power $= \frac{1}{f_{eq}} = 10 \, \mathrm{D}$

Given $V_1 = \frac{400}{100+400} \times 12 \, \mathrm{V} = \frac{4}{5} \times 12 = \frac{48}{5} \, \mathrm{V}$. Here, $V_1 > V_z$, ($V_z = Zener Voltage$). So, Zener breakdown will take place. Therefore, the voltage across $400 \, \Omega$ will be $4 \, \mathrm{V}$. The current $I = \frac{4}{400} \, \mathrm{A} = \frac{1}{100} \, \mathrm{A} = 10 \, \mathrm{mA}$.

Given $\theta_1 = 45 + \alpha$ and $\theta_2 = 45 - \alpha$. The time of flight $T$ is given by $$T = \frac{2v \sin \theta}{g}$$ For $T_1$: $$T_1 = \sin(45 + \alpha)$$ For $T_2$: $$T_2 = \sin(45 - \alpha)$$ Simplifying further, $$T_1 = \frac{1}{\sqrt{2}} \cos \alpha + \frac{1}{\sqrt{2}} \sin \alpha$$ $$T_2 = \frac{1}{\sqrt{2}} \cos \alpha - \frac{1}{\sqrt{2}} \sin \alpha$$ Therefore, $$T_1 = \frac{\cos \alpha + \sin \alpha}{\cos \alpha - \sin \alpha} = \frac{1 + \tan \alpha}{1 - \tan \alpha}$$

Given $\gamma = 1 + \frac{2}{f}$. (1) $f = 6$, Triatomic rigid gas $f = 7$, Diatomic non-rigid gas $f = 5$, Diatomic rigid gas $f = 3$, monoatomic rigid gas $\gamma = 1 + \frac{2}{6} = \frac{4}{3}$ (Triatomic) $\gamma = 1 + \frac{2}{7} = \frac{9}{7}$ (Diatomic, non-rigid) $\gamma = 1 + \frac{2}{5} = \frac{7}{5}$ (Diatomic, rigid) $\gamma = 1 + \frac{2}{3} = \frac{5}{3}$ (Monoatomic, rigid) A-III, B-IV, C-I, D-II

Given the forces acting on the block, we have the equation: $$mg \sin 60^\circ - \mu mg \cos 60^\circ = ma$$ Substituting the given acceleration $a = \frac{g}{2}$, we get: $$g \sin 60^\circ - \mu g \cos 60^\circ = \frac{g}{2}$$ Solving for $\mu$, we have: $$\frac{\sqrt{3}}{2} - \frac{\mu}{2} = \frac{1}{2}$$ Therefore, $$\mu = \sqrt{3} - 1$$

Given $\Delta E = 13.6 z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$. $$ (13.6) z^2 \left[ \frac{1}{1^2} - \frac{1}{9} \right] = 108.8 $$ $$ \frac{(13.6)(8)}{9} (z^2) = 108.8 $$ $$ z = 3 $$

For the Lyman series, when $n = 2$ to $n = 1$, we have: $$\frac{1}{\lambda_1} = R \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = \frac{3R}{4}$$ Thus, $$\lambda_1 = \frac{4}{3R} ...(1)$$ For the Balmer series, when $n = 3$ to $n = 2$, we have: $$\frac{1}{\lambda_2} = R \left[ \frac{1}{2^2} - \frac{1}{3^2} \right] = \frac{5R}{36}$$ Thus, $$\lambda_2 = \frac{36}{5R}$$ Then, $$\frac{\lambda_1}{\lambda_2} = \frac{5}{27}$$

Given the equations of motion in a magnetic field, we have: $$\frac{mv^2}{r} = qvB$$ Solving for $mv$, we get: $$mv = qBr$$ The kinetic energy $E$ is given by: $$E = \frac{1}{2} mv^2$$ Substituting $mv = qBr$ into the expression for $E$: $$E = \frac{1}{2} m \left( \frac{q^2 B^2 r^2}{m^2} \right) = \frac{q^2 B^2 r^2}{2m}$$ This simplifies to: $$E = \left( \frac{q^2 B^2}{2m} \right) r^2$$ The graph shows that $r^2 \propto E$.

Given $\vec{F} = (2t \hat{i} + 3t \hat{j}) \mathrm{N}$. $m = 1000 \mathrm{gm} = 1 \mathrm{kg}$. $\vec{F} = m \vec{a}$, $\vec{a} = 2t \hat{i} + 3t^2 \hat{j}$. $\frac{d \vec{v}}{dt} = 2t \hat{i} + 3t^2 \hat{j}$. $\vec{v} = t^2 \hat{i} + t^3 \hat{j}$. Power, $P = \vec{F} \cdot \vec{v}$. $$P = (2t \hat{i} + 3t^2 \hat{j}) \cdot (t^2 \hat{i} + t^3 \hat{j})$$ $$P = (2t^3 + 3t^5) \mathrm{W}$$

Given $\frac{L_A}{L_B} = \frac{1}{3}$ and $\frac{d_A}{d_B} = 2$. $$\frac{\Delta L_A}{\Delta L_B} = \frac{F_A L_A}{A_A Y_A} and \frac{\Delta L_B}{\Delta L_B} = \frac{F_B L_B}{A_B Y_B}$$ Given, $F_A = F_B$ and $Y_A = Y_B$. $$\frac{\Delta L_A}{\Delta L_B} = \frac{F_A L_A}{A_A Y_A} = \left(\frac{L_A}{L_B}\right) \left(\frac{A_B}{A_A}\right)$$ $$\frac{\Delta L_A}{\Delta L_B} = \left(\frac{L_A}{L_B}\right) \left(\frac{\frac{\pi}{4} d_B^2}{\frac{\pi}{4} d_A^2}\right) = \left(\frac{L_A}{L_B}\right) \left(\frac{d_B}{d_A}\right)^2$$ $$\frac{\Delta L_A}{\Delta L_B} = \left(\frac{1}{3}\right) \left(\frac{1}{2}\right)^2 = \frac{1}{12}$$

Potential at C $$V_C = \frac{kq_1}{0.4} + \frac{kq_2}{0.5}$$ Potential at D $$V_D = \frac{kq_1}{0.4} + \frac{kq_2}{0.1}$$ $$\Delta U = (V_D - V_C)(q_3) = \left(\frac{kq_2}{0.1} - \frac{kq_2}{0.5}\right)(q_3)$$ $$\Delta U = 8kq_2q_3 = \frac{8q_2q_3}{4\pi \varepsilon_0}$$

All bodies have same mass and same radius. A $\rightarrow$ Disc B $\rightarrow$ Solid sphere C $\rightarrow$ Spherical shell and, I = $\frac{MR^2}{4}$ I_{PQ} = $\frac{MR^2}{4}$ + $\left$( $\frac{2}{5}$ MR^2 + MR^2 $\right$) + $\left$( $\frac{2}{3}$ MR^2 + MR^2 $\right$) I_{PQ} = $\frac{15MR^2 + 24MR^2 + 60MR^2 + 40MR^2 + 60MR^2}{60}$ I_{PQ} = $\frac{199}{60}$ MR^2 = $\frac{199}{15}$ $\left$( $\frac{MR^2}{4}$ $\right$) = $\frac{199}{15}$ I

Given the input voltage, we have the following relationships: $$\theta + \theta = 90^\circ; \theta = 45^\circ$$ The tangent of the angle is given by: $$\tan \theta = \frac{X_C}{R}$$ Since $X_C = R$, it follows that: $$\frac{1}{W_C} = R$$ The work done $W$ is: $$W = \frac{1}{R_C} = \frac{10^{12}}{10^7} = 10^5$$

Let $y$ be the apparent depth of the bottom. $$y = \frac{H}{1.6} + \frac{60}{1.2}$$ Shift $= 40$ $$H + 60 - y = 40$$ Substituting for $y$: $$H + 60 - \frac{H}{1.6} - \frac{60}{1.2} = 40$$ Simplifying: $$\frac{6}{16} H = 30$$ Solving for $H$: $$H = 80 \, cm$$

Given $L = 10 \, \mathrm{cm}$, $d = 0.5 \, \mathrm{mm}$, $T = 1727^\circ \mathrm{C} = 2000 \, \mathrm{K}$. Power, $P = 94.2 \, \mathrm{W}$. $P = \varepsilon \sigma A T^4$ $$94.2 = \varepsilon \times (6 \times 10^{-8}) (\pi d L) (2000)^4$$ $$94.2 = \varepsilon \times (6 \times 10^{-8}) (3.14)(0.5) (10^{-3}) (10 \times 10^{-2}) (2000)^4$$ $$\varepsilon = \frac{94.2}{(94.2)(16)} = \frac{5}{8}$$

Area of circle, $W = \frac{\pi}{4} d_1 d_2$. $$W = \frac{\pi}{4} (500 - 300) \times 10^3 (350 - 150) \times 10^{-6}$$ $W = 31.4$ Joule $W = 314 \times 10^{-1}$ Joule

St.-I: Cis-2-butene is treated with (i) $\mathrm{O_3/Zn}$ and (ii) $\mathrm{H_2O}$ to form $2[\mathrm{CH_3-CHO}]$, which is ethanol. St.-II: The compound $\mathrm{H_3C-CH_2-CH-CH=CH-CH_2-CH_3}$ with $\mathrm{CH_3}$ groups is treated with (i) $\mathrm{O_3/Zn}$ and (ii) $\mathrm{H_2O}$ to form $2[\mathrm{CH_3-CH_2-CH-CO-CH_3}]$. The product has a chiral center. St-I: Correct statement St-II: Incorrect statement because product has chiral centre.

Only $1^\circ$ or primary amines gives positive carbylamines test. $$\mathrm{R{-}NH_2 \xrightarrow{CHCl_3 + KOH} R{-}N \equiv C}$$ (Alkyl iso cyanide, pungent) Option (A) and (C) are primary amine and give positive carbylamine test.

The reaction is given as $\mathrm{A(g)} \rightarrow 2\, \mathrm{B(g)} + \mathrm{C(g)}$. At $t = 0$, the pressure is $P_0$. As $t \rightarrow \infty$, the pressures are $0$, $2P_0$, and $P_0$ respectively. The total pressure at infinity is $P_\infty = 3P_0 = 240$. Therefore, $P_0 = 80 \, \mathrm{mm \, of \, Hg}$. The equation for $Kt$ is $$Kt = \ln \left( \frac{P_\infty - P_0}{P_\infty - P_t} \right).$$ Substituting the values, we have $$K \times 10 = \ln \left( \frac{240 - 80}{240 - 160} \right).$$ Solving this gives $$K = \frac{\ln 2}{10} = 0.0693 \, \mathrm{min^{-1}}.$$ Therefore, option (3) is incorrect.

The process involves the conversion of $\mathrm{H_2O_{(l)}}$ at $10^\circ \mathrm{C}$ to $\mathrm{H_2O_{(s)}}$ at $10^\circ \mathrm{C}$, and then to $\mathrm{H_2O_{(s)}}$ at $0^\circ \mathrm{C}$. The enthalpy change is calculated as follows: $$\Delta H = 1 \times y(0 - 10) - x \times 1000 + 1 \times z(-10^\circ - 0^\circ)$$ Simplifying, we have: $$\Delta H = -10(100x + y + z) Joule.$$

HCl(aq) pH = 1; $[\mathrm{H}^+] = 10^{-1}$. If equal volume of water is added, concentration will become half. $$[\mathrm{H}^+]_{sol} = \frac{10^{-1}}{2}$$ pH = 1.3

St-I is correct because both given ether are soluble in water. Diethyl ether and butan-1-ol are miscible to almost the same extent, i.e., 7.5 and 9 gm per 100 ml water due to H-bonding. St-II is also correct because sodium metal is not used with ethyl alcohol as $\mathrm{H_2}$ gas is released with ethyl alcohol.

Given $$\lambda = \frac{C}{\nu} = \frac{3 \times 10^8}{5.16 \times 10^{14}}$$ $$\lambda = 581.39 \, \mathrm{nm}$$ * $\lambda_{Photon}$ is near & below yellow light it can show photoelectric effect. * If intensity of light decreases photocurrent decreases. * Red light will not produce photoelectric effect. * $\nu_{Blue} > \nu_{yellow}$ so photoelectric current will be produced. * White light contain all frequencies so it will show photo electric current. Correct statements are A B D & E.

The compound shown is 1-Bromo-2-methyl but-2-ene.

Using Dalton's law of partial pressure, we have: $$\frac{P_{\mathrm{N_2}}}{P_{\mathrm{O_2}}} = \frac{x_{\mathrm{N_2}} \cdot P_T}{x_{\mathrm{O_2}} \cdot P_T} = \frac{n_{\mathrm{N_2}}}{n_{\mathrm{O_2}}}$$ Calculating the ratio: $$\frac{70/28}{27/32} = 2.96$$ For $P_{\mathrm{O_2}}$ and $P_{\mathrm{Ar}}$, we have: $$P_{\mathrm{O_2}} = n_{\mathrm{O_2}} = \frac{27/32}{3/40} = 11.25$$

Mohr's salt: $\mathrm{FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O}$ Using Kohlrausch law $$\lambda_m^\infty (Mohr's salt) = x_1 + 2x_2 + 2x_3$$

Out of Cr, Co, Fe and Ni, chromium has the lowest heat of atomisation. $$Cr = [Ar] 3d^5 4s^1$$ Therefore, total six valence electrons in Cr.

$NH_4^{+}+NaOH\xrightarrow{H_2O}NH_3\uparrow$

As per given information of ionisation energy, A = $\mathrm{Sn}$ $\&$ B = $\mathrm{Pb}$. $A^{+2}$ = $\mathrm{Sn}^{2+}$ = Reducing agent. $B^{+4}$ = $\mathrm{Pb}^{+4}$ = Oxidising agent.

In 3d series, Vanadium has the highest enthalpy of atomization and the color of $\mathrm{V^{+3}}$ is green. Oxide formed by $\mathrm{V^{+3}}$ is $\mathrm{V_2O_3}$ (Basic oxide).

Concept - Those compounds which are more acidic than $\mathrm{H_2CO_3}$ can give effervescence of $\mathrm{CO_2}$ with aqueous $\mathrm{NaHCO_3}$. Release $\mathrm{CO_2}$ gas with aqueous $\mathrm{NaHCO_3}$. $$[A.S.]_{Comp.} > [A.S.]_{\mathrm{H_2CO_3}}$$ Option 1, 2, and 3 give effervescence of $\mathrm{CO_2}$ gas with $\mathrm{NaHCO_3}$. Option (4) does not give $\mathrm{CO_2}$ gas with $\mathrm{NaHCO_3}$.

Before applying medicine $$\frac{dA}{dt} = K[A]$$ (First order growth) (Rate law) $$\frac{A}{A_0} = \frac{N}{N_0} = e^{Kt}$$ After applying medicine Active Bacteria $\rightarrow$ Inactive Bacteria $$r = -\frac{dA}{dt} = K[A]^2$$ (Rate law) $$y = Kx^2$$ Parabola

On hydrolysis of sucrose gives D-(+)-glucose and D-(-)-fructose while in St. (1) D-(+)-fructose is given evince. St-(1) is incorrect. St. II - It is correct because sucrose on hydrolysis gives invert sugar.

(A) complex is $[\mathrm{Co(NH_3)_5(SO_4)}] \mathrm{Cl}$ (3) (B) complex is $[\mathrm{Co(NH_3)_5Cl}] \mathrm{SO_4}$ Both (A) and (B) are Ionisation isomers.

Option (B) and (D) reaction are not able to form alkene as a product.

Organic compound $\xrightarrow{CuO} CO_2 + H_2O$ $n_{CO_2} = \frac{220 \times 10^{-3}}{44} = 5 \times 10^{-3} moles$ $m_C = 5 \times 10^{-3} \times 12$ $\% m carbon = \frac{5 \times 10^{-3} \times 12}{500 \times 10^{-3}} \times 100 = 12\%$ Correct answer is 12

Molecular formula of Thyroxine $\Rightarrow \mathrm{C_{15}H_{11}O_4N_1I_4}$. Molecular mass of Thyroxine: C $\rightarrow 15 \times 12 = 180$ H $\rightarrow 11 \times 1 = 11$ O $\rightarrow 16 \times 4 = 64$ N $\rightarrow 14 \times 1 = 14$ I $\rightarrow 127 \times 4 = 508$ Molecular mass of Thyroxine $\Rightarrow 777$. Percentage of Iodine $= \frac{508}{777} \times 100 = 65.38\%$. Nearest integer $= 65$.

The reaction is $\mathrm{Cu^{+2} + 2e^- \rightarrow Cu}$. One faraday is the charge on 1 mole of electrons. Initially, at $t = 0$, there are 1.5 moles of $\mathrm{Cu^{+2}}$. At $t = t$, there is 1 mole, so 0.5 moles have reacted. The concentration of $\mathrm{Cu^{+2}}$ is 1M after electrolysis. The reaction for silver is $\mathrm{Ag^+ + e^- \rightarrow Ag}$. Initially, at $t = 0$, there are 0.2 moles of $\mathrm{Ag^+}$. At $t = t$, there are 0.1 moles remaining. The concentration of $\mathrm{Ag^+}$ is 0.1M after electrolysis. The cell reaction is $\mathrm{Cu_{(s)} + 2Ag^+_{(aq)} \rightarrow Cu^{+2}_{(aq)} + 2Ag_{(s)}}$. The Nernst equation is given by: $$E = E^\circ - \frac{0.06}{n} \log \left( \frac{[\mathrm{Cu^{+2}}]}{[\mathrm{Ag^+}]^2} \right)$$ Substituting the values: $$E = (0.8 - 0.34) - \frac{0.06}{2} \log \left( \frac{1}{(0.1)^2} \right) = 0.4 \, \mathrm{V}$$ The correct answer is 400mV.

The reaction is given by $\($ $\mathrm{MX_3}$ $\rightarrow$ $\mathrm{M^{+3}}$ + 3$\mathrm{X^{\ominus}}$ $\)$. The formula for $\($ i $\)$ is $\($ i = 1 + (n - 1)$\alpha$ $\)$. Substituting the values, $\($ i = 1 + (4 - 1)$\alpha$ = 2 $\)$. Solving for $\($ $\alpha$ $\)$, we get $\($ $\alpha$ = $\frac{1}{3}$ = 33.33$\%$ $\approx$ 33$\%$ $\)$.

Q12. $[\mathrm{FeF}_6]^{3-}$ Paramagnetic $sp^3d^2$ $[\mathrm{Fe(CN)}_6]^{3-}$ Paramagnetic $d^2sp^3$ $[\mathrm{Mn(CN)}_6]^{3-}$ Paramagnetic $d^2sp^3$ $[\mathrm{Co(C_2O_4)}_3]^{3-}$ Diamagnetic $d^2sp^3$ $[\mathrm{MnCl}_6]^{3-}$ Paramagnetic $sp^3d^2$ $[\mathrm{CoF}_6]^{3-}$ Paramagnetic $sp^3d^2$ Only $[\mathrm{Fe(CN)}_6]^{3-}$ and $[\mathrm{Mn(CN)}_6]^{3-}$ are paramagnetic and $d^2sp^3$ hybridisation of metal.