1
MCQ (Single Correct Answer)

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 + $${1 \over 3}$$ $${{Mg} \over {\pi {R^2}Y}}$$)
2
MCQ (Single Correct Answer)

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 = $$\alpha $$
3
MCQ (Single Correct Answer)

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$$^{ - {{kt} \over m}}$$
4
MCQ (Single Correct Answer)

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 $${\left( {{H \over {H + g}}} \right)^{{1 \over 4}}}$$

Questions Asked from Properties of Matter

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