## IIT JEE Main 2016 Physics Question No 14

A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals µ . The particle is released, from rest, from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR. The values of the coefficient of friction µ and the distance x(= QR), are respectively close to

(1) 0.29 and 3.5 m
(2) 0.29 and 6.5 m
(3) 0.2 and 6.5 m
(4) 0.2 and 3.5 m

#### Solution

PQ = (h/sin 30º) = 2h = 4m
Given that work done against friction along track PQ and QR are equal.
therefore, µ mg cos 30º  x 4 =  µ mg x
(4√3)/2=x
=> x = 2√3 = 2 x 1.732 = 3.46 ≅ 3.5 m
Loss in PE = work done against friction along PQ and QR

hence, mg 2 = µ mg( (√3)/2)x4 + µ mg (3.5)

So we get µ = 0.29

Correct answer to given problem No 14 of IIT JEE Main Physics paper of 2016 is µ = 0.29 and QR = 3.5 i.e. option number 1.

## 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