Category Archives: IIT JEE

IIT JEE

2-D Collision

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A ball of mass 10 kg moving with a velocity $10\sqrt 3$ m/s along x-axis, hits another ball of mass 20 kg which is at rest kept at the origin. After collision, the first ball comes to rest and the second one disintegrates into two pieces. One of the pieces moves along y-axis at a speed of 10m/s. The second piece of mass m (say) moves at a speed of 20m/s at an angle $\theta$ (degree) with respect to the x–axis. Find the values of m and $\theta $.

Solution

Using conservation of linear momentum (COLM) along x direction,

$10\times 10 \sqrt 3 + 20 \times 0 = 10 \times 0 + m \times 20 \cos \theta + (20-m) \times 0$ Continue reading 2-D Collision →

IIT JEE

System of Masses $m_1 + m_2 + m_3 $

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Three masses $m_1 , m_2 , m_3 $ kept on a smooth horizontal surface under the influence of force F have got certain acceleration (refer figure). Find the force that mass $m_1$ exerts on mass $m_2$.

Solution

Let a be the acceleration.

Taking $m_1 + m_2 + m_3 $ as system,

$F = (m_1 + m_2 + m_3)a$

$\Rightarrow a = \frac {F}{m_1 + m_2 + m_3 }$

Let N be the force that $m_1 $ exerts on $m_2 $. Taking $m_2 + m_3$ as system,

$N = (m_2 + m_3 ) a$

$\Rightarrow N = F. \frac {m_2 + m_3 }{m_1 + m_2 + m_3 } $

Interestingly, as $m_1 \to 0$, $N \to F$

IIT JEE

Work Done in Hanging a Chain

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A chain with increasing mass per unit length from one end as a function of distance x from lighter end as $\lambda (x) = kx$ lying horizontally is to be hung against a vertical wall in such a manner that the external agent has to do minimum work. The min. work done by external agent is given by,

(A) $\frac {1}{6} kgl^3 $
(B) $\frac {1}{3} kgl^3 $
(C) $\frac {1}{2} kgl^3 $
(D) $\frac {2}{3} kgl^3 $

Solution

A vertically hanging chain with heavy end upwards has much more potential energy in comparison to other situations. If the heavy end is downwards, the potential energy is minimum. Minimum work would be required if increase in potential energy is minimum.

If horizontal plane is taken as the reference level for potential energy, then the initial potential energy is 0.

Let us calculate the final potential energy of vertical chain with heavy end at the bottom just touching the horizontal plane. Continue reading Work Done in Hanging a Chain →

IIT JEE

$3x+4y=10$

$x>0,y>0$

$(x^2y^3)_{max}=?$

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For positive numbers, A.M. $\geq $ G.M.

So, $\frac{{\frac{{3x}}{2} + \frac{{3x}}{2} + \frac{{4y}}{3} + \frac{{4y}}{3} + \frac{{4y}}{3}}}{5} \ge {\left[ {{{\left( {\frac{{3x}}{2}} \right)}^2}{{\left( {\frac{{4y}}{3}} \right)}^3}} \right]^{1/5}}$

$ \Rightarrow \frac{{3x + 4y}}{5} \ge {\left( {\frac{9}{4}{x^2} \times \frac{{4 \times 16}}{{9 \times 3}}{y^3}} \right)^{1/5}}$

$ \Rightarrow \frac{{10}}{5} \ge {\left( {\frac{{16}}{3}{x^2}{y^3}} \right)^{1/5}}$

$ \Rightarrow {2^5} \ge \frac{{16}}{3}{x^2}{y^3}$

$ \Rightarrow {x^2}{y^3} \le 6$

$ \Rightarrow {\left( {{x^2}{y^3}} \right)_{\max }} = 6$

IIT JEE

${x^2}f(x) + f\left( {\frac{1}{x}} \right) = 0,x \ne 0$

$I = \int\limits_{\tan \alpha }^{\cot \alpha } {f(x)dx} = ?$

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We have, $f(x) = – \frac{1}{{{x^2}}}f\left( {\frac{1}{x}} \right)$

$\therefore I = \int\limits_{\tan \alpha }^{\cot \alpha } { – \frac{1}{{{x^2}}}f\left( {\frac{1}{x}} \right)dx} $

Let, $\frac{1}{x} = t$

$ \Rightarrow – \frac{1}{{{x^2}}}dx = dt$

$\therefore I = \int\limits_{\cot \alpha }^{\tan \alpha } {f(t)dt} =-I $

$ \Rightarrow I = 0$

IIT JEE

Find $f(x)$ such that,

$f(x) + f\left( {\frac{1}{x}} \right) + 3f( – x) = x$

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$x \to \frac{1}{x}$ in the original equation yields,

$f\left( {\frac{1}{x}} \right) + f(x) + 3f\left( { – \frac{1}{x}} \right) = \frac{1}{x}$ ……..(A)

Original equation – (A) yields,

$3f( – x) – 3f\left( { – \frac{1}{x}} \right) = x – \frac{1}{x}$ Continue reading Find $f(x)$ such that,

$f(x) + f\left( {\frac{1}{x}} \right) + 3f( – x) = x$ →

IIT JEE

Let $I = \int\limits_0^{\pi /2} {\frac{{dx}}{{1 + {{\tan }^{\sqrt {\tan \alpha } }}x}}}$

$\frac{{dI}}{{d\alpha }} = ?$

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We have, $I = \int\limits_0^{\pi /2} {\frac{{dx}}{{1 + {{\tan }^{\sqrt {\tan \alpha } }}(\pi /2 – x)}}} $

$ \Rightarrow I = \int\limits_0^{\pi /2} {\frac{{dx}}{{1 + {{\cot }^{\sqrt {\tan \alpha } }}x}}} = \int\limits_0^{\pi /2} {\frac{{{{\sin }^{\sqrt {\tan \alpha } }}x.dx}}{{{{\sin }^{\sqrt {\tan \alpha } }}x + {{\cos }^{\sqrt {\tan \alpha } }}x}}} $ ……..(A)

Also, $I = \int\limits_0^{\pi /2} {\frac{{dx}}{{1 + {{\tan }^{\sqrt {\tan \alpha } }}x}} = } \int\limits_0^{\pi /2} {\frac{{{{\cos }^{\sqrt {\tan \alpha } }}x.dx}}{{{{\cos }^{\sqrt {\tan \alpha } }}x + {{\sin }^{\sqrt {\tan \alpha } }}x}}}$ ……..(B)

(A) + (B) gives, $2I = \int\limits_0^{\pi /2} {dx} = \frac{\pi }{2}$

$ \Rightarrow I = \frac{\pi }{4}$

$\therefore \frac{{dI}}{{d\alpha }} = 0$