1
JEE Main 2016 (Offline)
MCQ (Single Correct Answer)
+4
-1
Change Language
An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity $$C$$ remains constant. If during this process the relation of pressure $$P$$ and volume $$V$$ is given by $$P{V^n} = $$ constant, then $$n$$ is given by (Here $${C_p}$$ and $${C_v}$$ are molar specific heat at constant pressure and constant volume, respectively:
A
$$n = {{{C_p} - C} \over {C - {C_v}}}$$
B
$$n = {{C - {C_v}} \over {C - {C_p}}}$$
C
$$n = {{{C_p}} \over {{C_v}}}$$
D
$$n = {{C - {C_p}} \over {C - {C_v}}}$$
2
JEE Main 2016 (Offline)
MCQ (Single Correct Answer)
+4
-1
Change Language
A pendulum clock loses $$12$$ $$s$$ a day if the temperature is $${40^ \circ }C$$ and gains $$4$$ $$s$$ a day if the temperature is $${20^ \circ }C.$$ The temperature at which the clock will show correct time, and the co-efficient of linear expansion $$\left( \alpha \right)$$ of the metal of the pendulum shaft are respectively :
A
$${30^ \circ }C;\,\,\alpha = 1.85 \times {10^{ - 3}}/{}^ \circ C$$
B
$${55^ \circ }C;\,\,\alpha = 1.85 \times {10^{ - 2}}/{}^ \circ C$$
C
$${25^ \circ }C;\,\,\alpha = 1.85 \times {10^{ - 5}}/{}^ \circ C$$
D
$${60^ \circ }C;\,\,\alpha = 1.85 \times {10^{ - 4}}/{}^ \circ C$$
3
JEE Main 2016 (Offline)
MCQ (Single Correct Answer)
+4
-1
Change Language
A roller is made by joining together two cones at their vertices $$0$$. It is kept on two rails $$AB$$ and $$CD$$, which are placed asymmetrically (see figure), with its axis perpendicular to $$CD$$ and its center $$O$$ at the center of line joining $$AB$$ and $$CD$$ (see figure). It is given a light push so that it starts rolling with its center $$O$$ moving parallel to $$CD$$ in the direction shown. As it moves, the roller will tend to : JEE Main 2016 (Offline) Physics - Rotational Motion Question 236 English
A
go straight
B
turn left and right alternately
C
turn left
D
turn right
4
JEE Main 2016 (Offline)
MCQ (Single Correct Answer)
+4
-1
Change Language
A person trying to lose weight by burning fat lifts a mass of $$10$$ $$kg$$ upto a height of $$1$$ $$m$$ $$1000$$ times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies $$3.8 \times {10^7}J$$ of energy per $$kg$$ which is converted to mechanical energy with a $$20\% $$ efficiency rate. Take $$g = 9.8\,m{s^{ - 2}}$$ :
A
$$9.89 \times {10^{ - 3}}\,\,kg$$
B
$$12.89 \times {10^{ - 3}}\,kg$$
C
$$2.45 \times {10^{ - 3}}\,\,kg$$
D
$$6.45 \times {10^{ - 3}}\,\,kg$$

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