JEE Main 24 January 2025 Shift 1 question paper with solutions

Given $\vec{c} = \lambda (\vec{b} \times (\vec{a} \times \vec{b}))$. This equals $\lambda ((\vec{b} \cdot \vec{b}) \vec{a} - (\vec{a} \cdot \vec{b}) \vec{b})$. Simplifying, we have $\lambda (11 \vec{a} - 2 \vec{b}) = \lambda (11i + 22j + 33k - 6i - 2j + 2k)$. This results in $\lambda (5i + 20j + 35k)$. Thus, $= 5\lambda (5i + 4j + 7k)$. Given $\vec{c} \cdot \vec{a} = 5$, we have $5\lambda (1 + 8 + 21) = 5$ which implies $\lambda = \frac{1}{30}$. Therefore, $\Rightarrow \vec{c} = \frac{1}{6} (i + 4j + 7k)$. The magnitude of $\vec{c}$ is $\vec{c} = \frac{\sqrt{1 + 16 + 49}}{6} = \frac{\sqrt{11}}{6}$.

Given $$I(m, m) = \int_0^1 x^{m-1}(1-x)^{n-1} \, dx$$ Let $x = \sin^2 \theta$, $dx = 2 \sin \theta \cos \theta \, d\theta$. Then $$I(m, n) = 2 \int_0^{\pi/2} (\sin \theta)^{2m-1} (\cos \theta)^{2n-1} \, d\theta$$ Now, $$I(9, 14) + I(10, 13) = 2 \int_0^{\pi/2} (\sin \theta)^{17} (\cos \theta)^{27} \, d\theta$$ $$+ 2 \int_0^{\pi/2} (\sin \theta)^{19} (\cos \theta)^{25} \, d\theta$$ $$= 2 \int_0^{\pi/2} (\sin \theta)^{17} (\cos \theta)^{25} \, d\theta$$ $$= I(9, 13)$$

Given the equation $f(x) - 6f\left(\frac{1}{x}\right) = \frac{35}{3x} - \frac{5}{2}$ $\dots$ (1). Multiply equation (1) by 6: $$6\left(f\left(\frac{1}{x}\right) - 6f(x)\right) = \frac{35x}{3} - \frac{5}{2}$$ This gives: $$6f\left(\frac{1}{x}\right) - 36f(x) = \frac{210x}{3} - \frac{30}{2} \dots (2)$$ Adding equations (1) and (2): $$-35f(x) = \frac{35}{3}\left[x + 6x\right] - \frac{5}{2}(1 + 6)$$ Simplifying gives: $$-f(x) = \frac{1}{3}\left(x + 6x\right) - \frac{1}{2}$$ Thus, $$f(x) = -\frac{1}{3x} - 2x + \frac{1}{2}$$ Taking the limit as $x \to 0$: $$\lim_{x \to 0}\left[\frac{1}{\alpha x} - \frac{1}{3x} - 2x + \frac{1}{2}\right] = \beta$$ This implies $\alpha = 3$. Calculating $\beta$: $$\beta + 2\beta = 3 + 2 \times \frac{1}{2} = 4$$ Therefore, $\beta = 4$.

Given the series $S_n = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \ldots$ for $N$ terms. $S_{2025} = \sum_{n=1}^{2025} \frac{1}{n(n+1)} = \sum_{n=1}^{2025} \left( \frac{1}{n} - \frac{1}{n+1} \right)$ $$= \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \ldots + \left( \frac{1}{2025} - \frac{1}{2026} \right)$$ $$= 2025$$ $$= 2026$$ $\sqrt{2026} \cdot S_{2025} = \sqrt{2025} = 45$ Given: $\binom{6}{2} [-2p + (6-1)p] = 45$ $$9p = 45$$ $$p = 5$$ $|A_{20} - A_{15}| = |-5 + 19 \times 5| - [-5 + 14 \times 5]$ $$= |90 - 65|$$ $$= 25$$

Given $$f(x) = \frac{4^{2x} + 16}{2 \cdot 2^{2x} + 16 \cdot 2^x + 32}$$ Simplifying, $$f(x) = \frac{2^{2x} + 8 \cdot 2^x + 16}{2(2^x + 4)}$$ Further simplifying, $$f(x) = \frac{2^x}{2(2^x + 4)}$$ Now, $$f(4-x) = \frac{2^x}{2(2^x + 4)}$$ Thus, $$f(x) + f(4-x) = \frac{1}{2}$$ So, $$f\left(\frac{1}{15}\right) + f\left(\frac{59}{15}\right) = \frac{1}{2}$$ Similarly, $$f\left(\frac{29}{15}\right) + f\left(\frac{31}{15}\right) = 1$$ Now, $$f\left(\frac{30}{15}\right) = f(2) = \frac{2}{2^2 + 4} = \frac{2}{8} = \frac{1}{4}$$ Therefore, $$\Rightarrow 8\left(29 \times \frac{1}{2} + \frac{1}{4}\right)$$ Ans. 118

Given the equation $2z^2 - 3z - 2i = 0$ (i). (1) $2 \left( z - \frac{i}{z} \right) = 3$ As $\alpha$, $\beta$ are roots of (i) $$\alpha - \frac{i}{\alpha} = \frac{3}{2}$$ $$\Rightarrow \alpha^2 - \frac{1}{\alpha} - 2i = \frac{9}{4}$$ $$\Rightarrow \alpha^2 - \frac{1}{\alpha^2} = \frac{9}{4} + 2i$$ Squaring both sides $$\Rightarrow \alpha^4 + \frac{1}{\alpha^4} - 2 = \frac{81}{16} - 4 + 9i$$ $$\Rightarrow \alpha^4 + \frac{1}{\alpha^4} = \frac{49}{16} + 9i$$ Similarly, $\beta^4 + \frac{1}{\beta^4} = \frac{49}{16} + 9i$ $$\alpha^{19} + \beta^{19} + \alpha^{11} + \beta^{11} = \alpha^{15} + \beta^{15}$$ $$= \alpha^{15} \left( \alpha^4 + \frac{1}{\alpha^4} \right) + \beta^{15} \left( \beta^4 + \frac{1}{\beta^4} \right)$$ $$= \alpha^{15} + \beta^{15}$$ $$= \frac{49}{16} + 9i$$ $$Re \left( \frac{\alpha^{19} + \beta^{19} + \alpha^{11} + \beta^{11}}{\alpha^{15} + \beta^{15}} \right) = \frac{49}{16}$$ $$Im \left( \frac{\alpha^{19} + \beta^{19} + \alpha^{11} + \beta^{11}}{\alpha^{15} + \beta^{15}} \right) = 9$$ $$\Rightarrow 16 \cdot Re \left( \frac{\alpha^{19} + \beta^{19} + \alpha^{11} + \beta^{11}}{\alpha^{15} + \beta^{15}} \right) \cdot Im \left( \frac{\alpha^{19} + \beta^{19} + \alpha^{11} + \beta^{11}}{\alpha^{15} + \beta^{15}} \right)$$ $$= 16 \times \frac{49}{16} \times 9$$ $$= 441$$

Given the limit expression: $$ \lim_{x \to 0} \csc x \left( \sqrt{2 \cos^2 x + 3 \cos x - \sqrt{\cos^2 x + \sin x + 4}} \right) $$ We start by simplifying the expression inside the cosecant function: $$ \cos ec \, x \left( \cos^2 x + 3 \cos x - \sin x - 4 \right) $$ Taking the limit as $x$ approaches 0: $$ \lim_{x \to 0} \left( \sqrt{2 \cos^2 x + 3 \cos x + \sqrt{\cos^2 x + \sin x + 4}} \right) $$ Simplifying further: $$ \lim_{x \to 0} \frac{1}{\sin x} \left( \sqrt{2 \cos^2 x + 3 \cos x + \sqrt{\cos^2 x + \sin x + 4}} \right) $$ Using the identity $\sin x \approx x$ as $x \to 0$: $$ \lim_{x \to 0} \sin x \left( \sqrt{2 \cos^2 x + 3 \cos x + \sqrt{\cos^2 x + \sin x + 4}} \right) $$ Simplifying the expression: $$ \lim_{x \to 0} \sin x \left( \cos^x + 4 \right) (\cos x - 1) - \sin x $$ Further simplification gives: $$ \lim_{x \to 0} \sin x \left( \sqrt{2 \cos^2 x + 3 \cos x + \sqrt{\cos^2 x + \sin x + 4}} \right) $$ Simplifying the trigonometric terms: $$ -2 \sin^2 \frac{x}{2} (\cos x + 4) - 2 \sin \frac{x}{2} \cos \frac{x}{2} $$ Taking the limit: $$ \lim_{x \to 0} 2 \sin \frac{x}{2} \cos \frac{x}{2} \left( \sqrt{2 \cos^2 x + 3 \cos x + \sqrt{\cos^2 x + \sin x + 4}} \right) $$ Simplifying further: $$ - \left( \sin^x (\cos x + 4) + \cos \frac{x}{2} \right) $$ Finally, we have: $$ \lim_{x \to 0} \cos \frac{x}{2} \left( \sqrt{2 \cos^2 x + 3 \cos x + \sqrt{\cos^2 x + \sin x + 4}} \right) $$ The final result is: $$ - \frac{1}{2 \sqrt{5}} $$

Let M(3$\lambda$ + 6, 2$\lambda$ + 7, -2$\lambda$ + 7). $$\overrightarrow{\mathrm{BM}} = (3\lambda + 5)\hat{i} + (2\lambda + 5)\hat{j} + (-2\lambda + 4)\hat{k}$$ $$\overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{BM}} = 0 = 3(3\lambda + 5) + 2(2\lambda + 5) - 2(-2\lambda + 4)$$ $$\overrightarrow{\mathrm{BM}} = 2\hat{i} + 3\hat{j} + 6\hat{k}$$ $$|\overrightarrow{\mathrm{BM}}| = 7$$ $$Area = \frac{1}{2} \times 6 \times 7 = 21$$

Given $\($(1 + x^2) $\frac{dy}{dx}$ + xy = 5x^2 $\sqrt{1 + x^2}$$\)$. Rewriting, we have $\($$\frac{dy}{dx}$ + $\frac{xy}{1 + x^2}$ = $\frac{5x^2}{\sqrt{1 + x^2}}$$\)$. The integrating factor is $\($$\mathrm{I.F.}$ = e^{$\int$ $\frac{x}{1+x^2}$ $\,$ dx} = e^{$\frac{\ln(1+x^2)}{2}$} = $\sqrt{1 + x^2}$$\)$. Thus, $\($y $\sqrt{1 + x^2}$ = $\int$ $\frac{5x^2}{\sqrt{1 + x^2}}$ $\cdot$ $\sqrt{1 + x^2}$ $\,$ dx$\)$. This simplifies to $\($y $\sqrt{1 + x^2}$ = $\int$ 5x^2 $\,$ dx$\)$. Integrating, we get $\($y $\sqrt{1 + x^2}$ = $\frac{5x^3}{3}$ + C$\)$. Given $\($y(0) = 0$\)$, we find $\($0 = 0 + C $\Rightarrow$ C = 0$\)$. Thus, $\($y = $\frac{5x^3}{3\sqrt{1 + x^2}}$$\)$. Finally, $\($y($\sqrt{3}$) = $\frac{15\sqrt{3}}{32}$ = $\frac{5\sqrt{3}}{2}$$\)$.

Product of focal distances is $(a + ex_1)(a - ex_1)$. (2) $= a^2 - e^2x_1^2 = a^2 - e^2(3)$ $= a^2 - 3e^2 = \frac{7}{4} \Rightarrow a^2 = \frac{7}{4} + 3e^2$ $\Rightarrow 4a^2 = 7 + 12e^2$ Lines on $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $\therefore \frac{a^2}{3} + \frac{4b^2}{1} = 1$ $\frac{3}{a^2} + \frac{1}{4(a^2)(1-e^2)} = 1$ $12(1-e^2) + 1 = 4a^2(1-e^2)$ $13 - 12e^2 = (7 + 12e^2)(1-e^2)$ $\Rightarrow 13 - 12e^2 = 7 - 7e^2 + 12e^2 - 12e^4$ $\Rightarrow 12e^4 - 17e^2 + 6 = 0$ $\therefore e^2 = \frac{17 \pm \sqrt{289 - 288}}{24} = \frac{17 \pm 1}{24} = \frac{3}{4} \& \frac{2}{3}$ $\therefore e = \frac{\sqrt{3}}{2} \& \sqrt{\frac{2}{3}}$ $\therefore difference == \frac{\sqrt{3}}{2} - \sqrt{\frac{2}{3}} = \frac{3 - 2\sqrt{2}}{2\sqrt{3}}$

For sum '5' $\rightarrow$ (1,4), (2,3), (3,2) (4,1) $\Rightarrow$ P(A) = $\frac{4}{36}$ For sum '8' $\rightarrow$ (2,6), (3,5), (4,4) For sum '5' $\rightarrow$ (1,4), (2,3), (3,2) (4,1) $\Rightarrow$ P(A) = $\frac{4}{36}$ For sum '8' $\rightarrow$ (2,6), (3,5), (4,4) (5,3), (6,2) $\Rightarrow$ P(B) = $\frac{5}{36}$ $$P(\bar{A}) = \frac{32}{36}, P(\bar{B}) = \frac{31}{36}$$ $$P(A wins ) = P(A) + P(\bar{A})P(\bar{B})P(A) + P(\bar{A})P(\bar{B})P(\bar{A})P(\bar{B})P(A) + \ldots$$ $$= \frac{P(A)}{1 - P(\bar{A})P(\bar{B})} = \frac{9}{19}$$

Given the area function $A = 9t - t^2 - \frac{11}{3} t^3$. Differentiate with respect to $t$ to find $\frac{dA}{dt} = 9 - 2t - 11t^2$. Setting $\frac{dA}{dt} = 0$ gives $11t^2 + 2t - 9 = 0$. Solving this quadratic equation, $11t^2 + 11t - 9 = 0$ gives $t = \frac{-1 \pm \sqrt{81}}{11}$. The maximum occurs at $t = \frac{9}{11}$. Therefore, the largest area is $\frac{9}{11} \left( 9 - \frac{11}{3} \cdot \frac{81}{121} - \frac{9}{11} \right) = \frac{9}{11} \cdot \frac{63}{11} = \frac{567}{121}$.

Given the inequality $x^2 + 4x + 2 \leq y \leq |x + 2|$. The required area $A_1$ is calculated as follows: $$A_1 = \int_{-4}^{0} \left[ 2 - (x^2 + 4x + 2) \right] \, dx - \frac{1}{2} \times 4 \times 2$$ Evaluating the integral: $$= \left( \frac{-x^3}{3} - 2x^2 \right)_{-4}^{0} - 4$$ $$= \left( 0 - \left( \frac{64}{3} - 32 \right) \right) - 4$$ $$= 32 - \frac{64}{3} - 4 = \frac{20}{3}$$

Mean $\bar{x} = 5.5$ $$\sum_{i=1}^{10} x_i = 5.5 \times 10 = 55$$ $$\sum_{i=1}^{10} x_i^2 = 371$$ $$(\sum x_i)_{new} = 55 - (4 + 5) + (6 + 8) = 60$$ $$(\sum x_i^2)_{new} = 371 - (4^2 + 5^2) + (6^2 + 8^2) = 430$$ Variance $\sigma^2 = \frac{\sum x_i^2}{10} - \left(\frac{\sum x_i}{10}\right)^2$ $$\sigma^2 = \frac{430}{10} - \left(\frac{60}{10}\right)^2$$ $$\sigma^2 = 43 - 36$$ $$\sigma^2 = 7$$

Centre $(1, -2)$, $r = 3$. Reflection of $(1, -2)$ about $2x - 3y + 5 = 0$. $$\frac{x - 1}{2} = \frac{y + 2}{-3} = \frac{-2(2 + 6 + 5)}{13} = -2$$ $$x = -3, \; y = 4$$ Equation of circle 'C' $$C : (x + 3)^2 + (y - 4)^2 = 9$$ A.T.Q. $$\ell(arc AB) = \frac{1}{6} \times 2\pi$$ $$r \theta = \frac{1}{6} \times 2\pi$$ $$\theta = \frac{\pi}{3}$$ $$(\alpha + 6)^2 + (\beta - 4)^2 = 27$$ $$(\alpha + 3)^2 + (\beta - 4)^2 = 9$$ $$(\alpha + 6)^2 - (\alpha + 3)^2 = 18$$ $$\Rightarrow 6\alpha = -9$$ $$\Rightarrow \alpha = -\frac{3}{2}, \; \beta = \left(4 - \frac{3\sqrt{3}}{2}\right)$$ $$\therefore \beta - \frac{\sqrt{3}\alpha}{\left(4 - \frac{3\sqrt{3}}{2}\right) + \frac{3\sqrt{3}}{2}} = 4$$

Given $(1+x)^{n+4}$. $n+4C_4$, $n+4C_5$, $n+4C_6$ form an A.P. Therefore, $$2 \times n+4C_5 = n+4C_4 + n+4C_6$$ $$4 \times n+4C_5 = (n+4C_4 + n+4C_5) + (n+4C_5 + n+4C_6)$$ $$4 \times n+4C_5 = n+5C_5 + n+5C_6$$ $$\frac{4 \times 5! \cdot (n-1)!}{(n+4)!} = \frac{6! \cdot n!}{(n+6)!}$$ $$\Rightarrow 4 = \frac{(n+6)(n+5)}{6n}$$ $$\Rightarrow n^2 + 11n + 30 = 24n$$ $$\Rightarrow n^2 - 13n + 30 = 0$$ $$\Rightarrow n = 3, 10 (rejected)$$ Therefore, $n \neq 10$. Thus, the largest binomial coefficient in the expansion of $(1+x)^7$ (since $n+4 = 7$) is the coefficient of the middle term. Therefore, $$7C_4 = 7C_3 = 35$$

Given $\left( x^2 - 9x + 11 \right)^2 - (x-4)(x-5) = 3$. Let $x^2 - 9x + 11 = t$. Then, $t^2 - (t + 9) = 3$. This implies $t^2 - t - 12 = 0$. Further, $t^2 - 4t + 3t - 12 = 0$. Factoring gives $t(t - 4) + 3(t - 4) = 0$. Therefore, $t = 4$ or $-3$. For $x^2 - 9x + 11 = 4$, we have $x^2 - 9x + 7 = 0$. Here, we will get irrational roots. For $x^2 - 9x + 11 = -3$, we have $x^2 - 9x + 14 = 0$. Solving gives $x^2 - 7x - 2x + 14 = 0$, which implies $x = 7, 2$. Therefore, the product of all rational roots is $14$.

The equation of the line through the point $(-1, 2, 1)$ is given by $$\frac{x+1}{2} = \frac{y-2}{3} = \frac{z-1}{4} = \lambda.$$ So, $$x = 2\lambda - 1,$$ $$y = 3\lambda + 2,$$ $$z = 4\lambda + 1.$$ By equation (1), $$\frac{x+2}{3} = \frac{y-3}{2} = \frac{z-4}{1} = \mu (Let).$$ So, $$x = 3\mu - 2,$$ $$y = 2\mu + 3,$$ $$z = \mu + 4.$$ For the intersection point $P$, $$x = 2\lambda - 1 = 3\mu - 2,$$ $$y = 3\lambda + 2 = 2\mu + 3,$$ $$z = 4\lambda + 1 = \mu + 4.$$ So, point $P(x, y, z) = (1, 5, 5)$. Let $Q(4, -5, 1)$. Therefore, $$PQ = \sqrt{9 + 100 + 16} = \sqrt{125} = 5\sqrt{5}.$$

Since $\overline{\mathrm{PM}} = \overline{\mathrm{QM}}$, we have: $$\mathrm{M} \left( \frac{57}{13} + 1 \cdot \frac{-40}{13} , \frac{1 + 2}{2} \right)$$ So, $\mathrm{M} \left( \frac{35}{13}, \frac{-7}{13} \right)$. Therefore, $\mathrm{M}$ lies on the line: $$2x - 3y + \lambda = 0$$ Substituting $\left( \frac{35}{13}, \frac{-7}{13} \right)$: $$2 \left( \frac{35}{13} \right) - 3 \left( \frac{-7}{13} \right) + \lambda = 0$$ $$\lambda = -\frac{70}{13} + \frac{21}{13}$$ $$= \frac{-91}{13} = -7$$ The matrix equation is: $$\begin{bmatrix} 3 & -4 & -\alpha \\ 8 & -11 & -33 \\ 2 & 3 & \lambda \end{bmatrix}$$ Solving: $$3(-11\alpha - 99) + 4(8\lambda + 66) - \alpha(-24 + 22) = 0$$ $$33\lambda - 297 + 32\lambda + 264 + 24\alpha - 22\alpha = 0$$ $$-\lambda + 2\alpha - 33 = 0$$ Thus, $\lambda = -7$. $$-(-7) + 2\alpha - 33 = 0$$ $$2\alpha = 26$$ $$\alpha = 13$$ Therefore, $|\alpha \lambda| = 13 \times (-7)| = 91$$

Given the equations: $$2x - y + z = 4$$ $$5x + \lambda y + 3z = 12$$ $$100x - 47y + \mu z = 212$$ We have $\Delta x = \Delta y = \Delta z = 0$. For $\Delta z$: $$\begin{vmatrix} 2 & -1 & 4 \\ 5 & \lambda & 12 \\ 100 & -47 & 212 \end{vmatrix} = 0$$ This implies: $$2(212\lambda + 564) + 1(1060 - 1200)$$ $$\Rightarrow 424\lambda + 1128 - 140 - 940 - 400\lambda = 0$$ $$\Rightarrow \lambda = -2$$ For $\Delta y$: $$\begin{vmatrix} 2 & -1 & 1 \\ 5 & -2 & 3 \\ 100 & -47 & \mu \end{vmatrix} = 0$$ This implies: $$2(-2\mu + 141) + 1(5\mu - 300) + 1(-235 + 200) = 0$$ $$\Rightarrow \mu = 53$$ Finally, $$\mu - 2\lambda = 53 - 2(-2) = 57$$

Differentiating both sides $$4(x+2)f(x) + 2f(x)(x+2)^2 - 6(x+2) = 10(x+2)f(x)$$ $$= (x+2) \frac{dy}{dx} - 3y = 3$$ $$\frac{1}{3} \int \frac{dy}{y+1} = \int \frac{dx}{x+2}$$ $$\ln |y+1| = 3 \ln |x+2| + \ln c$$ $$y+1 = (x+2)^3 c$$ Therefore, $$y(0) = \frac{3}{2}$$ $$\Rightarrow \frac{5}{16} = c$$ Thus, $$y = \frac{5}{16} (x+2)^3 - 1$$ $$y(2) = \frac{5}{16} \times 64 - 1 = 19$$

Let $\tan^{-1} \alpha = A \Rightarrow \tan A = \alpha$. $\cot^{-1} \beta = B \Rightarrow \cot B = \beta$. $\sec^2 A + \csc^2 B = 36$. $$\Rightarrow 1 + \tan^2 A + 1 + \cot^2 B = 36$$ $$\Rightarrow \alpha^2 + \beta^2 = 34$$ Also $\alpha + \beta = 8$ (Given). Therefore, $(\alpha + \beta)^2 = 34 + 2\alpha\beta = 64$. $$\Rightarrow \alpha\beta = 15$$ Therefore, $\alpha, \beta$ are roots of the equation $$x^2 - 8x + 15 = 0$$ $$\Rightarrow (x - 3)(x - 5) = 0$$ $$\Rightarrow x = 3, 5$$ Therefore, $\alpha = 3, \beta = 5 (\alpha < \beta)$. Therefore, $\alpha^2 + \beta = 9 + 5 = 14$.

No. of 3 digits = 999 - 99 = 900 No. of 3 digit numbers divisible by 2 & 3 i.e. by 6 $$\frac{900}{6} = 150$$ No. of 3 digit numbers divisible by 4 & 9 i.e. by 36 $$\frac{900}{36} = 25$$ Therefore, No. of 3 digit numbers divisible by 2 & 3 but not by 4 & 9 $$150 - 25 = 125$$

Given $X^T A X = 0$. $\begin{pmatrix} x & y & z \end{pmatrix} \begin{pmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = 0$ $\begin{pmatrix} x & y & z \end{pmatrix} \begin{pmatrix} a_1 x + a_2 y + a_3 z \\ b_1 x + b_2 y + b_3 z \\ c_1 x + c_2 y + c_3 z \end{pmatrix} = 0$ $x (a_1 x + a_2 y + a_3 z) + y (b_1 x + b_2 y + b_3 z) + z (c_1 x + c_2 y + c_3 z)$ $a_1 = 0, b_2 = 0, c_3 = 0$ $a_2 + b_1 = 0, a_3 + c_1 = 0, b_3 + c_2 = 0$ $A = skew symmetric matrix$ $A = \begin{pmatrix} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{pmatrix}; A \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \\ -5 \end{pmatrix}$ $\Rightarrow \begin{pmatrix} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \\ -5 \end{pmatrix}$ $x + y = 1$ $-x + z = 4$ $y + z = 5$ $\begin{pmatrix} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \\ -8 \end{pmatrix}$ $2x + y = 0 x = -1$ $-x + z = 4 y = 2$ $-y - 2z = -8 z = 3$ $A = \begin{pmatrix} 0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0 \end{pmatrix}$ $2(A + I) = \begin{pmatrix} 2 & -2 & 4 \\ 2 & 2 & 6 \\ -2 & -6 & 2 \end{pmatrix}$ $2(A + I) = 120$ $\Rightarrow \det(adj(2A + I)) = 120^2 = 2^6 \cdot 3^2 \cdot 5^2$ $\therefore \alpha = 6, \beta = 2, \gamma = 2$ Hence $\alpha^2 + \beta^2 + \gamma^2 = 6^2 + 2^2 + 2^2 = 44$

Let $\frac{y}{x} = \lambda$. Then $y = \lambda x$. $$y = 10 \times \left( \binom{9}{0} + \binom{9}{1} + \binom{9}{2} + \binom{9}{3} + \ldots + \binom{9}{9} \right)$$ $$= 10 \times 2$$ $$= 10 \times 512 = 5120$$

We know energy density $$\rho_{av} = \frac{1}{2} \varepsilon_0 E^2$$ So potential energy = $$\rho_{av} \times Volume$$ $$\Rightarrow P.E. = \frac{1}{2} \varepsilon_0 E^2 A d$$

Initial optical power, $P_1 = 2.5 \, \mathrm{D}$ Final optical power, $P_2 = 2.5 \, \mathrm{D} + 0.1 \, \mathrm{D} = 2.6 \, \mathrm{D}$ Step 1: Relation Between Focal Length and Power The focal length $f$ (in meters) of a lens is related to its optical power $P$ by: $$P = \frac{1}{f}$$ So, $$f_1 = \frac{1}{P_1} = \frac{1}{2.5} = 0.4 \, \mathrm{m}$$ $$f_2 = \frac{1}{P_2} = \frac{1}{2.6} \approx 0.3846 \, \mathrm{m}$$ Step 2: Relative Decrease in Focal Length The relative decrease in focal length is given by: $$\frac{\Delta f}{f_1} = \frac{f_1 - f_2}{f_1}$$ Substituting the values: $$\Delta f = \frac{0.4 - 0.3846}{0.4}$$ $$= \frac{0.0154}{0.4}$$ $$= 0.0385 which is approx 0.04$$ Final Answer: 0.04

Given $P_1 = P_0 + \rho gh = P_0 + 1000 \times 10 \times \frac{20}{100}$. Therefore, $P_1 = P_0 + 2000$. So, $P_2 - P_1 = \frac{2S}{R} = \left( \frac{2S}{1 \times 10^{-3}} \right)$. Thus, $P_2 = P_0 + 2100$. Given, So, $P_0 + 2100 - P_0 - 2000 = 2S$. Therefore, $100 = 2S \times 10^3$. Thus, $s = \left( \frac{1}{20} \right) = 0.05$.

Given $\($ y = $\frac{32.3 \times 1125}{27.4}$ = 1326.18 $\)$. So we need to report to three significant digits. So, $\($ y = 1330 $\)$.

For A to C $$\frac{hc}{\lambda_1} = E_0 z^2 \left( \frac{1}{n_c^2} - \frac{1}{n_A^2} \right) \ldots (i)$$ And $$\frac{hc}{\lambda_2} = E_0 z^2 \left( \frac{1}{n_c^2} - \frac{1}{n_B^2} \right)$$ So for A and B $\ldots$ (ii) $$\frac{hc}{\lambda_3} = E_0 z^2 \left( \frac{1}{n_B^2} - \frac{1}{n_A^2} \right)$$ Clearly subtracting equation (ii) from equation (i) $$hc \left[ \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right] = E_0 z^2 \left[ \frac{1}{n_B^2} - \frac{1}{n_A^2} \right] = \frac{hc}{\lambda_3}$$ $$\Rightarrow \frac{1}{\lambda_3} = \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \Rightarrow \frac{1}{\lambda_3} = \frac{(6000 - 2000)}{6000 \times 2000} = \frac{1}{3000}$$ $$\lambda_3 = 3000 \AA$$

Clearly statement 'A' is wrong as for solar cell junction area are mode wide. LED connected in forward bias and its intensity increases up to certain value of current and further there is no change as intensity saturates. Also solar cells are not connected with any external bias.

Given $W_1 = s 4 \pi \left( \frac{R}{3} \right)^2 \times 27 - 4 \pi R^2 \cdot s$. Therefore, $W_1 = 4 \pi R^2 \cdot s(2) = 10 Joule$. Now, $W_2 = s 4 \pi R^2 (4 - 1) = 4 \pi R^2 s \times 3$. Therefore, $W_2 = 3 \times \frac{10}{2} = 15 Joule$.

Given the initial velocity $v_0$ at an angle of $45^\circ$, the maximum height $H$ is given by: $$H = \left( \frac{v_0}{\sqrt{2}} \right)^2 \frac{1}{2g} = \frac{v_0^2}{4g}.$$ The angular momentum $L$ is given by $L = mvh$. Substituting the values, we have: $$L = m \frac{v_0}{\sqrt{2}} \frac{v_0^2}{4g}.$$

Given the equation $$\frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d}$$ and the conditions $$n_{480} = m_{600}$$ and $$n_{\min} = 5$$.

Given maximum possible speed $= v_0$. For this, So, $N = mg \cos \theta + \frac{mv_0^2}{r} \sin \theta$ $f_r = \mu mg \cos \theta + \mu \frac{mv_0^2}{r} \sin \theta$ And $$\frac{mv_0^2}{r} \cos \theta = mg \sin \theta + f_r$$ $$\Rightarrow \frac{mv_0^2}{r} \cos \theta - mg \sin \theta = \mu \left( mg \cos \theta + \frac{mv_0^2}{r} \sin \theta \right)$$ $$\Rightarrow (v_0^2 - gr \tan \theta) = \mu (v_0^2 \tan \theta + gr)$$ $$\Rightarrow \mu = \frac{v_0^2 - gr \tan \theta}{gr + v_0^2 \tan \theta}$$

For pure rolling about point $P$. $$\Rightarrow a = \frac{2g}{3} \sin \theta = \frac{2g}{3\sqrt{2}} = \frac{\sqrt{2}}{3} g$$

Given the equation $$\frac{1.5}{v} = \frac{1.5 - 1.2}{R}$$ we have $$v = \frac{1.5R}{0.3} = 5R.$$ Then, $$\frac{1.2}{f} - \frac{1.5}{5R} = \frac{1.2 - 1.5}{-R}.$$ Simplifying, $$\frac{1.2}{f} = \frac{0.3}{R}.$$ Therefore, $$f \times 2 = f = 2R \Rightarrow R = 0.1.$$

Given $A = 1 \, \mathrm{cm}$. $$n = \frac{12.5}{2} = 6.25 cycles$$ Therefore, $$D = 4 \times 6 + 1 = 25$$ $d = 1$ $$\frac{D}{d} = 25$$

Given $T_1 \propto (R)^{3/2}$ and $T_2 \propto (1.03R)^{3/2}$. Therefore, $$T_2 = (1.03R)^{3/2} \cdot T_1 \approx 1.045T_1.$$ So $T_2$ will be larger by 4.5$\%$ with respect to $T_1$.

Both lenses are plano-convex, so the second surface is plane ($R = \infty$), which simplifies the lens maker's formula: $$\frac{1}{f} = \left( n_n - 1 \right) \left( \frac{1}{R} - 0 \right)$$ where: - $n$ is the refractive index of the lens material, - $n_n$ is the refractive index of the surrounding medium, - $R$ is the radius of curvature of the curved surface. Step 1: Calculate $f_1$ (Lens in Air) For Lens 1 in Air: $$\frac{1}{f_1} = \left( \frac{1.5}{1} - 1 \right) \left( \frac{1}{2} \right)$$ $$\frac{1}{f_1} = (1.5 - 1) \times \frac{1}{2}$$ $$\frac{1}{f_1} = 0.5 \times \frac{1}{2} = \frac{0.5}{2} = \frac{1}{4}$$ $$f_1 = 4 \, \mathrm{cm}$$ Step 2: Calculate $f_2$ (Lens in Liquid) For Lens 2 in Liquid: $$\frac{1}{T_2} = \left( \frac{1.5}{1.2} - 1 \right) \left( \frac{1}{3} \right)$$ $$\frac{1}{T_2} = \left( \frac{1.5 - 1.2}{1.2} \right) \times \frac{1}{3}$$ $$\frac{1}{T_2} = \left( \frac{0.3}{1.2} \right) \times \frac{1}{3}$$ $$\frac{1}{T_2} = \frac{0.3}{3.6}$$ $$f_2 = \frac{3.6}{1.3} = 12 \, \mathrm{cm}$$ Step 3: Find the Ratio $\frac{4}{12}$ $$\frac{f_1}{f_2} = \frac{4}{12} = \frac{1}{3}$$ Final Answer: $$\frac{1}{3}$$

Given $i = i_1 \sin \omega t + i_2 \sin(\omega t + 90)$. We have $i = \sqrt{i_1^2 + i_2^2} \sin(\omega t + \phi)$. The rms current $i_{rms}$ is given by $$i_{rms} = \frac{i_0}{\sqrt{2}}$$ where $$i_0 = \sqrt{i_1^2 + i_2^2}$$ Thus, $$i_{rms} = \frac{\sqrt{i_1^2 + i_2^2}}{\sqrt{2}}$$

Given $\vec{v} = v_0 \hat{i} - \frac{E_0 e}{m} t \hat{k}$. The magnitude of $\vec{v}$ is given by $$|\vec{v}| = \sqrt{v_0^2 + \frac{E_0^2 e^2 t^2}{m^2}}.$$ The initial wavelength $\lambda_0$ is $$\lambda_0 = \frac{h}{mv_0}.$$ The new wavelength $\lambda'$ is $$\lambda' = \frac{h}{mv_0 \sqrt{1 + \frac{E_0^2 e^2 t^2}{v_0^2 m^2}}}.$$ Thus, $$\lambda' = \frac{\lambda_0}{\sqrt{1 + \frac{E_0^2 e^2 t^2}{v_0^2 m^2}}}.$$

We know $C = \frac{A \varepsilon_0}{d} = \frac{b \ell \varepsilon_0}{d}$. So to increase the capacitance by 10 factor $\left( \frac{A}{d} \right)$ has to increase by 10 factor. For option (A) $C' = \frac{(30 \ell) b \varepsilon_0}{\left( \frac{d}{3} \right)} = 30C$. For option (B) $C' = \frac{\ell b \varepsilon_0}{10d} = \frac{C}{10}$. For option (C) $C' = \frac{d}{(2 \ell) 5b \varepsilon_0} = 10C$. For option (D) $C' = \frac{\left( \frac{\ell}{3} \right) b \varepsilon_0}{\left( 10d \frac{3}{d} \right)} = \frac{C}{10}$. For option (E) $C' = \frac{\left( \frac{\ell}{3} \right) 5 (2b) \varepsilon_0}{\left( \frac{d}{3} \right)} = 10C$. Clearly (C) and (E) are the situation for 10C.

Given $F = \alpha + \beta x^2$. Work done $\int dw = \int F \cdot dx$. Therefore, $$\Delta W = \int F \cdot dx = \int (\alpha + \beta x^2) \, dx$$ $$\Rightarrow \Delta W = \left| \alpha x + \frac{\beta x^3}{3} \right|_0^1 = \alpha + \frac{\beta}{3} = 5$$ Given $\alpha = 1$. So, $$\frac{\beta}{3} = 4$$ Therefore, $$\beta = 12 \, \mathrm{N/m^2}$$

Given that $$P = kT$$ $$\frac{P}{T} = constant$$ Therefore, volume is constant or isochoric process. Thus, $W_D = 0$ and $Q = \Delta U$. Also, temperature increases hence internal energy increases.

Total flux through square = $\frac{q}{\varepsilon_0} \left( \frac{1}{6} \right)$. Let's divide the square into 8 equal parts. Flux is the same for each part. Therefore, flux through shaded portion is $\frac{5}{8}$ (Total flux) $$= \frac{5}{8} \times \frac{q}{\varepsilon_0} \frac{1}{6} = \frac{5}{48} \frac{1}{\varepsilon_0}$$ Therefore, required answer is 48.

Given least count of Screw Gauge = 0.01 $\mathrm{mm}$ $\newline$ $\mathrm{L.C}$ = $\frac{\text{pitch}}{\text{No. of circular turn}}$ = $\frac{P}{N}$ = 0.01 $\mathrm{mm}$ $\newline$ New pitch = $\frac{P(1 + 0.75)}{N(1 - 0.5)}$ = $\frac{P \left[ 1.75 \right]}{N \left[ 0.5 \right]}$ $\newline$ = (0.01) 3.5 $\newline$ = 0.035 $\mathrm{mm}$ $\newline$ = 35 $\times$ 10^{-3} $\mathrm{mm}$

Given that $9\,\Omega$ is the resistance of the whole wire. Therefore, the resistance of each wire is $3\,\Omega$. Thus, the equivalent resistance is $2\,\Omega$.

Let B be the magnetic field due to a single side. Then $B = \frac{\mu_0 i}{4 \pi \ d} (\sin \theta_1 + \sin \theta_2)$. $$= 10^{-7} \times 5 \times 2 \times \frac{1}{\sqrt{2}} = 2 \times 10^{-6}$$ Therefore, $B_{net}$ at centre $O = 4B$. $$= 8 \times 10^{-6}$$

Given that process is isobaric $\Delta T = 50^\circ \mathrm{C}$. Q in isobaric process $= nC_p \Delta T = E_1$. $\Delta U$ in isobaric process $= nC_v \Delta T = E_2$. Therefore, $\frac{E_1}{E_2} = \frac{C_p}{C_v} = \gamma$. Given, gas is monoatomic. Therefore, $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{3} = \frac{5}{3}$. Now, as per question, $$\frac{5}{3} = \frac{x}{9}$$ $$x = 15$$

Given the reaction $\mathrm{Fe^{2+}(aq) + Ag^+(aq) \rightarrow Fe^{3+}(aq) + Ag(s)}$. $$\Delta G_3^0 = \Delta G_1^0 + \Delta G_2^0$$ $$-3 \, F(-z) = -2 \, F(-y) + \Delta G_2^0$$ $$\Delta G_2^0 = 3Fz - 2Fy$$ Also, $$\Delta G_2^0 = -nF E^0_{\mathrm{Fe^{+2}/Fe^{t43}}}$$ $$3Fz - 2Fy = -1 \, F \left( E^0_{\mathrm{Fe^{+2}/Fe^{t43}}} \right)$$ $$E^0_{\mathrm{Fe^{+2}/Fe^{t3}}} = 2y - 3z$$ $E^0_{Cell}$ for reaction will be $$E^0_{\mathrm{Ag^+/Ag}} + E^{03}_{\mathrm{Fe^{t2}/Fe}}$$ $$= x + 2y - 3z$$

Addition of $\mathrm{H} - \mathrm{Br}_{(aq)}$ to alkene follows electrophilic addition mechanism. In the rate determining step a carbocation intermediate is formed. Among $\mathrm{P}$, $\mathrm{Q}$, $\mathrm{R}$ $\&$ $\mathrm{S}$ compound $\mathrm{Q}$ will form most stable carbocation intermediate since it is resonance stabilized.

The reaction is as follows: $$[\mathrm{Co(NH_3)_5Cl}] \mathrm{Cl_2} \rightarrow [\mathrm{Co(NH_3)_5Cl}]^{2+} \,(\mathrm{aq.}) + 2\mathrm{Cl}^- \,(\mathrm{aq.})$$ This results in 3 ions in water. The further reaction is: $$[\mathrm{Co(NH_3)_5Cl}] \mathrm{Cl_2} \,(\mathrm{aq.}) + 2\mathrm{AgNO_3} \,(\mathrm{aq.}) \rightarrow$$ $$[\mathrm{Co(NH_3)_5Cl}] \,(\mathrm{NO_3})_2 \,(\mathrm{aq.}) + 2\mathrm{AgCl(s)}$$

The compound $\overset{+}{\mathrm{CH_2}}- \mathrm{CH=CH-O-CH_3}$ will be most stable due to extended conjugation. The compound (3) will be less stable due to the $-M$ effect of $\overset{-}{\mathrm{C}}=\mathrm{O}$.

A: $p_x$ along $Z$ and $p_x$ along $X$ results in no molecular orbital. B: $s$ along $Z$ and $p_x$ along $X$ results in no molecular orbital. C: $d_{xy}$ along $Y$ and $d_{x^2-y^2}$ along $Y$ results in no molecular orbital. D: $s$ along $Z$ and $p_z$ along $Z$ results in a Sigma molecular orbital. E: $p_x$ along $Z$ and $d_{x^2-y^2}$ along $Y$ results in no molecular orbital.

Tb^{4+} is the strongest oxidising agent as it will reduce to Tb^{3+} (common O.S. of Ln).

The dissociation of $\mathrm{Cr(OH)_3}$ is given by: $$\mathrm{Cr(OH)_3 \rightleftharpoons Cr^{3+} (aq) + 3OH^- (aq)}$$ The solubility product is expressed as: $$K_{sp} = s(3s)^3$$ Simplifying, we have: $$1.6 \times 10^{-30} = 27s^4$$ Solving for $s$: $$\sqrt[4]{\frac{1.6 \times 10^{-30}}{27}} = s$$

On increasing temperature, the given endothermic reaction becomes spontaneous. $$\Delta G = \Delta H - T \Delta S$$ For spontaneity, $$\Delta G 0$$ So, $$\Delta S$$ should be positive.

In Dumas method nitrogen present in organic compound is converted into $\mathrm{N_2}$ gas whose volumetric analysis gives the percentage of nitrogen atom in the organic compound.

Properties of elements are function of atomic numbers. Elements having similar outer electronic configuration are arranged in same group. Number of elements in a period is double of number of atomic orbitals available in energy level that is being filled.

In Ribose carbohydrate present in DNA is $\beta - 2$-Deoxy-D-Ribose whose structure is which is a reducing $D$-sugar in $\beta$ anomeric form and it is a pentose sugar.

Preparation of $\mathrm{KMnO_4}$ $\mathrm{MnO_2} \xrightarrow[\mathrm{KNO_3}]{\mathrm{KOH}} K_2MnO_4 \xrightarrow[\text{Oxidation in alkaline solution}]{\text{Electrolysis}} KMnO_4$

The large difference in the melting and boiling points of oxygen and sulphur is due to atomicity as oxygen exists as $\mathrm{O_2}$ and sulphur exists as $\mathrm{S_8}$.

The reaction is given by $\mathrm{N_2O_5}_{(g)} \rightarrow 2\mathrm{NO_2}_{(g)} + \frac{1}{2}\mathrm{O_2}_{(g)}$. At $t = 0$, the pressure is $P_0$. At time $t = t$, the pressure of $\mathrm{N_2O_5}$ is $P_0 - x$, the pressure of $\mathrm{NO_2}$ is $2x$, and the pressure of $\mathrm{O_2}$ is $\frac{x}{2}$. Solving for $x$, we have $x = \frac{P_0}{2}$. The total pressure $P_{total}$ is given by: $$P_{total} = P_0 - \frac{P_0}{2} + P_0 + \frac{P_0}{4} = \frac{7}{4} P_0$$

Reactivity is proportional to the positive charge on electrophilic carbon. The correct order is I > II > III > IV.

On adding non-volatile solute in a solvent, the freezing point of solution decreases. $T_f < T_f^0$ F.P. of solution < F.P. of pure solvent. Also V.P. of solution decreases on adding nonvolatile solute in a solvent.

Dipole moment $\mathrm{H_2O} > \mathrm{NH_3} > \mathrm{CH_4}$. $\mathrm{H_2O}$ and $\mathrm{NH_3}$ are Lewis Bases. $\mathrm{NH_3}$ acts as a Lowry-Bronsted base. Hence, A, B & C are correct.

Initial pressure of $\mathrm{N_2O_5}$ is calculated as follows: $$\frac{37.8}{108} \times 0.082 \times 500 = 14.35 \, bar$$ The reaction is: $$2 \, \mathrm{N_2O_5} \rightleftharpoons 2 \, \mathrm{N_2O_4} + \mathrm{O_2}$$ At $t = 0$, the pressure is $14.35$. At equilibrium, $t = eq$, the pressures are $14.35 - 2P$ and $2P$. The total pressure at equilibrium is $\mathrm{P_{Total}} = 14.35 + P = 18.65$. Solving for $P$, we get $P = 4.3$. The partial pressures are: $$\mathrm{P_{N_2O_5}} = 5.75 \, bar$$ $$\mathrm{P_{N_2O_4}} = 8.6 \, bar$$ $$\mathrm{P_{O_2}} = 4.3 \, bar$$ The equilibrium constant $k_p$ is calculated as: $$k_p = \frac{(8.6)^2 \times (4.3)}{(5.75)^2} = 9.619 = x \times 10^{-2}$$ Thus, $x = 961.9 \approx 962$.

For the reaction $\frac{1}{2} X_2 + \frac{5}{2} Y_2 \rightarrow XY_5$, the change in entropy $\Delta S^0_{Rxn}$ is calculated as follows: $$\Delta S^0_{Rxn} = 110 - \left[ \left( \frac{1}{2} \times 70 \right) + \left( \frac{5}{2} \times 50 \right) \right]$$ This simplifies to: $$= 110 - 160 = -50 \, \mathrm{JK^{-1} \, mol^{-1}}$$ At equilibrium, $\Delta G^0 = 0$. Therefore, $$\Delta G^0 = \Delta H^0 - T \Delta S^0$$ Substituting the values: $$0 = -35000 - T(-50)$$ Solving for $T$ gives: $$T = 700 \, Kelvin$$

1 mole benzoic acid will release 1 mole $\mathrm{CO_2}$. 11.2 L $\mathrm{CO_2}$ at STP is $\frac{1}{2}$ mole $\mathrm{CO_2}$, which will be released by reaction with $\frac{1}{2}$ mole benzoic acid. Mass of benzoic acid $= \frac{1}{2} \times 122 = 61 \, \mathrm{gm}$.

Only $\mathrm{Cu^{2+}}$, $\mathrm{Fe^{3+}}$, $\mathrm{Ca^{2+}}$ and $\mathrm{Zn^{2+}}$ form precipitate with $\mathrm{K_4[Fe(CN)_6]}$.

moles of $\mathrm{Fe_3O_4} = \frac{2.32 \times 10^3}{232} = 10000 \, \mathrm{mol}$ moles of $\mathrm{CO} = \frac{2.8 \times 10^2 \times 10^3}{28} = 10000 \, \mathrm{mol}$ $$\mathrm{Fe_3O_4} + 4\mathrm{CO} \rightarrow 3\mathrm{Fe} + 4\mathrm{CO_2}$$ $10^4 \, \mathrm{mol} 10^4 \, \mathrm{mol}$ $\mathrm{CO}$ is L.R. mole of $\mathrm{Fe} = \frac{3}{4} \times 10^4$ mass of $\mathrm{Fe} = \frac{3}{4} \times \frac{10^4 \times 56}{1000} \, \mathrm{kg} = 420 \, \mathrm{kg}$