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### JEE Main 2016 (Online) 9th April Morning Slot

A uniformly tapering conical wire is made from a material of Young’s modulus Y and has a normal, unextended length L. The radii, at the upper and lower ends of this conical wire, have values R and 3 R, respectively. The upper end of the wire is fixed to a rigid support and a mass M is suspended from its lower end. The equilibrium extended length, of this wire, would equal :
A
L $\left( {1 + {2 \over 9}{{Mg} \over {\pi Y{R^2}}}} \right)$
B
L $\left( {1 + {1 \over 3}{{Mg} \over {\pi Y{R^2}}}} \right)$
C
L $\left( {1 + {1 \over 9}{{Mg} \over {\pi Y{R^2}}}} \right)$
D
L $\left( {1 + {2 \over 3}{{Mg} \over {\pi Y{R^2}}}} \right)$

## Explanation Here r = 3R $-$ ${{2R} \over L}$ x

$\therefore$   Extension in the wire of length dx,

dl = ${{Fdx} \over {AY}}$

= ${{Mg\,dx} \over {\pi {r^2}\,Y}}$

= ${{Mg\,dx} \over {\pi {{\left( {3R - {{2R} \over L}x} \right)}^2}Y}}$

$\therefore$   Change in wire length,

$\Delta$L = $\int\limits_0^L {dl}$

= $\int\limits_0^L {{{Mg\,dx} \over {\pi {{\left( {3R - {{2R} \over L}x} \right)}^2}Y}}}$

= ${{Mg} \over {\pi Y}}\int\limits_0^L {{{dx} \over {{{\left( {3R - {{2R} \over L}x} \right)}^2}}}}$

= ${{Mg} \over {\pi Y}}\left[ { - {1 \over {\left( {3R - {{2R} \over L}x} \right)}} \times \left( { - {L \over {2R}}} \right)} \right]_0^L$

= ${{Mg} \over {\pi Y}}$ $\left[ {\left( {{L \over {2{R^2}}} - {L \over {6{R^2}}}} \right)} \right]$

= ${{Mg} \over {\pi Y}}\left( {{{2L} \over {6{R^2}}}} \right)$

= ${{MgL} \over {3\pi {R^2}Y}}$

$\therefore$   The equilibrium extended length of the wire,

= L + $\Delta$L

= L + ${{MgL} \over {3\pi {R^2}Y}}$

= L (1 + ${1 \over 3}$ ${{Mg} \over {\pi {R^2}Y}}$)
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### JEE Main 2016 (Online) 9th April Morning Slot

A simple pendulum made of a bob of mass m and a metallic wire of negligible mass has time period 2 s at T=0oC. If the temperature of the wire is increased and the corresponding change in its time period is plotted against its temperature, the resulting graph is a line of slope S. If the coefficient of linear expansion of metal is $\alpha$ then the value of S is :
A
$\alpha$
B
${\alpha \over 2}$
C
2$\alpha$
D
${1 \over \alpha }$

## Explanation

Change of length of wire with temperature,

$\Delta $$\ell = \alpha \ell \Delta \theta Time period of pendulum at temperature \theta , T\theta = 2\pi$$\sqrt {{{\ell + \Delta \ell } \over g}}$

= 2$\pi$$\sqrt {{{\ell \left( {1 + \alpha \Delta \theta } \right)} \over g}}$

= $2\pi \sqrt {{\ell \over g}} {\left( {1 + \alpha \Delta \theta } \right)^{{1 \over 2}}}$

= $2\pi \sqrt {{\ell \over g}} {\left( {1 + {{\Delta \ell } \over \ell }} \right)^{{1 \over 2}}}$

$\simeq$  T0 $\left( {1 + {{\Delta \ell } \over {2\ell }}} \right)$

Here T0 = time period at temperature 0oC.

$\therefore$   Change in time period,

$\Delta$T = T$\theta$ $-$ T0

= ${{{T_0}\Delta \ell } \over {2\ell }}$

= ${{{T_0}\left( {\alpha \ell \Delta \theta } \right)} \over {2\ell }}$

$\therefore$   ${{\Delta T} \over {\Delta \theta }}$ = ${{{T_0}\alpha } \over 2}$

Given that T0 = 2,

$\therefore$   ${{\Delta T} \over {\Delta \theta }}$ = ${{2\alpha } \over 2}$ = $\alpha$

${{\Delta T} \over {\Delta \theta }}$ is the shape of $\Delta$T and ${\Delta \theta }$

curve = S (given)

$\therefore$   S = $\alpha$
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### JEE Main 2016 (Online) 9th April Morning Slot

Which of the following option correctly describes the variation of the speed v and acceleration ‘a’ of a point mass falling vertically in a viscous medium that applies a force F = − kv, where ‘k’ is a constant, on the body ? (Graphs are schematic and not drawn to scale)
A B C D ## Explanation

Equation of motion for the mass,

ma = mg $-$ kv

$\Rightarrow$    ${{dv} \over {dt}} = {{mg - kv} \over m}$

$\Rightarrow$   $\int\limits_0^v {{{dv} \over {mg - kv}}} = {1 \over m}\int\limits_0^t {dt}$

$\Rightarrow$   $- {1 \over k}\left[ {\ln \left( {mg - kv} \right)} \right]_0^v = {t \over m}$

$\Rightarrow$   $\ln \left( {{{mg - kv} \over {mg}}} \right) = - {{kt} \over m}$

$\Rightarrow$   $1 - {{kv} \over {mg}} = {e^{ - {{kt} \over m}}}$

$\Rightarrow$   ${{kv} \over {mg}} = 1 - {e^{ - {{kt} \over m}}}$

$\Rightarrow$   $v = {{mg} \over k}\left( {1 - {e^{ - {{kt} \over m}}}} \right)$

ma $=$ mg $-$ k $\times$ ${{mg} \over k}$ (1 $-$ e$^{ - {{kt} \over m}}$)

$=$  mg $-$ mg + mge$^{ - {{kt} \over m}}$

a $=$ g e$^{ - {{kt} \over m}}$
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### JEE Main 2016 (Online) 9th April Morning Slot Consider a water jar of radius R that has water filled up to height H and is kept on astand of height h (see figure). Through a hole of radius r (r << R) at its bottom, the water leaks out and the stream of water coming down towards the ground has a shape like a funnel as shown in the figure. If the radius of the cross-section of water stream when it hits the ground is x. Then :
A
$x = r\left( {{H \over {H + h}}} \right)$
B
$x = r{\left( {{H \over {H + h}}} \right)^{{1 \over 2}}}$
C
$x = r{\left( {{H \over {H + h}}} \right)^{{1 \over 4}}}$
D
$x = r{\left( {{H \over {H + h}}} \right)^{{2}}}$

## Explanation

v1 = velocity of water when it leak from hole

v2 = velocity of water when it reach the ground.

From Bernoulli's principle,

${1 \over 2}\rho {v_1}^2 + \rho gh$ = ${1 \over 2}\rho {v_2}^2$

$\Rightarrow$   ${v_1}^2$ + 2gh = ${v_2}^2$

From Torricelli's theorem,

v1 = $\sqrt {2gH}$

$\therefore$   ${v_2}^2$ = 2gh + 22gH

From continuity equation,

${A_1}{v_1}$ = ${A_2}{v_2}$

$\Rightarrow$   $\pi {r^2} \times \sqrt {2gH}$ = $\pi {x^2}\sqrt {2g\left( {h + H} \right)}$

$\Rightarrow$   x2  =  r2$\sqrt {{H \over {H + g}}}$

$\Rightarrow$   x = r ${\left( {{H \over {H + g}}} \right)^{{1 \over 4}}}$