IIT JEE Main 2016 Physics Question No 13

IIT JEE Main 2016 Physics Question No 13

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 × 107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. [Take g = 9.8 ms–2].
(1) 9.89 × 10–3 kg
(2) 12.89 × 10–3 kg
(3) 2.45 × 10–3 kg
(4) 6.45 × 10–3 kg

Solution

Total potential energy, U is given by,
U = mgh × 1000
=> U = 10 × 9.8 × 1 × 1000
Since 20% efficiency rate, hence
3.8 × 10000000 × 20/100 m = 9.8 X 10000
=> m = (9.8 × 10000)/(3.8 × 1000000 × 2)
=> m = 12.89/1000 kg
=> 12.89 × 10–3 kg

Answer

Correct answer to given problem No 13 of IIT JEE Main Physics paper of 2016 is 12.89 × 10–3 kg i.e. option number 2.

IIT JEE Main 2016 Physics Question No 12

IIT JEE Main 2016 Physics Question No 12

‘n’ moles of an ideal gas undergoes a process A->B as shown in the figure. The maximum temperature of the gas during the process will be:

IIT JEE Main 2016 Physics Question No 12

(1) 9P0V0/(2nR)
(2) 9P0V0/(nR)
(3) 9P0V0/(4nR)
(4) 3P0V0/(2nR)

Solution

The equation of the process is y = mx + c
P = -(P0/V0)V + 3P0
Now PV = nRT
therefore P = nRT/V
=> (nRT/V)=-(P0/V0)V + 3P0
=> nRT=-(P0/V0)V2 + 3P0V
=> T=-(P0/nRV0)V2 + (3P0/nR)V                  … (1)
Now for maximum T,
dT/dV = 0                                                       … (2)
here dT/dV = -(2P0/nRV0)V + (3P0/nR)       … (3)
from (2) & (3) we get,
V=(3/2)V0
putting in (1) we get
T=-(P0/nRV0)(9/4V02) + (3P0/nR)(3/2)V0
=> T=(9/2)(P0V0/nR)-(9/4)(P0V0/nR)
=> Tmax = (9/4)(P0V0/nR)

Answer

Correct answer to given problem No 12 of IIT JEE Main Physics paper of 2016 is Tmax = (9/4)(P0V0/nR) i.e. option number 3.