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1

AIEEE 2012

MCQ (Single Correct Answer)
A wooden wheel of radius $$R$$ is made of two semicircular part (see figure). The two parts are held together by a ring made of a metal strip of cross sectional area $$S$$ and length $$L.$$ $$L$$ is slightly less than $$2\pi R.$$ To fit the ring on the wheel, it is heated so that its temperature rises by $$\Delta T$$ and it just steps over the wheel. As it cools down to surrounding temperature, it process the semicircular parts together. If the coefficient of linear expansion of the metal is $$\alpha $$, and its Young's modulus is $$Y,$$ the force that one part of the wheel applies on the other part is :
A
$$2\pi SY\alpha \Delta T$$
B
$$SY\alpha \Delta T$$
C
$$\pi SY\alpha \Delta T$$
D
$$2SY\alpha \Delta T$$

Explanation

$$\gamma = {{F/S} \over {\Delta L/L}} \Rightarrow \Delta L = {{FL} \over {SY}}$$
$$\therefore$$ $$L\alpha \Delta T = {{FL} \over {SY}}$$
$$\left[ \, \right.$$ as $${\Delta L = L\alpha \Delta T}$$ $$\left. \, \right]$$
$$\therefore$$ $$F = SY\alpha \Delta T$$
$$\therefore$$ The ring is pressing the wheel from both sides,
$$\therefore$$ $${F_{net}} = 2F = 2YS\alpha \Delta T$$
2

AIEEE 2012

MCQ (Single Correct Answer)
A liquid in a beaker has temperature $$\theta \left( t \right)$$ at time $$t$$ and $${\theta _0}$$ is temperature of surroundings, then according to Newton's law of cooling the correct graph between $${\log _e}\left( {\theta - {\theta _0}} \right)$$ and $$t$$ is:
A
B
C
D

Explanation

Newton's law of cooling
$${{d\theta } \over {dt}} = - k\left( {\theta - {\theta _0}} \right)$$
$$ \Rightarrow {{d\theta } \over {\left( {\theta - {\theta _0}} \right)}} = - kdt$$

Intergrating
$$ \Rightarrow \log \left( {\theta - {\theta _0}} \right) = - kt + c$$
Which represents an equation of straight line.
Thus the option $$(a)$$ is correct.

3

AIEEE 2012

MCQ (Single Correct Answer)
Helium gas goes through a cycle $$ABCD$$ (consisting of two isochoric and isobaric lines) as shown in figure efficiency of this cycle is nearly : (Assume the gas to be close to ideal gas)
A
$$15.4\% $$
B
$$9.1\% $$
C
$$10.5\% $$
D
$$12.5\% $$

Explanation

Heat given to system $$ = {\left( {n{C_v}\Delta T} \right)_{A \to B}} + {\left( {n{C_p}\Delta T} \right)_{B \to C}}$$
$$ = {\left[ {{3 \over 2}\left( {nR\Delta T} \right)} \right]_{A \to B}} + {\left[ {{5 \over 2}\left( {nR\Delta T} \right)} \right]_{B \to C}}$$
$$ = {\left[ {{3 \over 2} \times {V_0}\Delta P} \right]_{A \to B}} + {\left[ {{5 \over 2} \times 2{P_0} \times {V_0}} \right]_{B \to C}}$$
$$ = {{13} \over 2}{P_0}{V_0}$$
and $${W_0} = {P_0}{V_0}$$
$$\eta = {{Work} \over {heat\,\,given}}$$
$$ = {{{P_0}{V_0}} \over {{{13} \over 2}{P_0}{V_0}}} \times 100$$
$$ = 15.4\% $$
4

AIEEE 2012

MCQ (Single Correct Answer)
A Carnot engine, whose efficiency is $$40\% $$, takes in heat from a source maintained at a temperature of $$500$$ $$K.$$ It is desired to have an engine of efficiency $$60\% .$$ Then, the intake temperature for the same exhaust (sink) temperature must be :
A
efficiency of Carnot engine cannot be made larger than $$50\% $$
B
$$1200$$ $$K$$
C
$$750$$ $$K$$
D
$$600$$ $$K$$

Explanation

$$0.4 = 1 - {{{T_2}} \over {500}}\,\,\,\,$$ and $$\,\,\,\,0.6 = 1 - {{{T_2}} \over {{T_1}}}$$
on solving we get $${T_2} = 750\,K$$

Questions Asked from Heat and Thermodynamics

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