## IIT JEE Main 2017 Physics Question 11 SetA:

IIT JEE Main 2017 Physics Question 11 SetA states that the temperature of an open room of volume 30 m3 increases from 17 deg C to 27 deg C due to the sunshine. The atmospheric pressure in the room remains 1 x 105 Pa. If ni and nf are the number of molecules in the room before and after heating, then nf – ni will be:

 (1) -1.61 x 1023 (2) 1.38 x 1023 (3) 2.5 x 1025 (4) -2.5 x 1025

#### Solution

According to the problem, there is constant atmospheric pressure in the room and:
n1 = initial number of moles
=> n1 = (P1V1)/(RT1)
=> n1 = (105 X 30)/(8.3 X 290)
=> n1 = 1.24 X 103

n2 = final number of moles
=> n2 = (P2V2)/(RT2)
=> n2 = (105 X 30)/(8.3 X 300)
=> n2 = 1.20 X 103

Change in number of molecules :
nf – ni = (n2 – n1) x 6.023 x 1023
nf – ni = -2.5 X 1025

#### Answer to IIT JEE Main 2017 Physics Question 11 SetA

Correct answer to given problem No 11 of IIT JEE Main Physics paper of 2017 Set A is option number (4) that is the Change in number of molecules is  nf – ni = -2.5 X 1025

## IIT JEE Main 2017 Physics Question 10 SetA:

IIT JEE Main 2017 Physics Question 10 SetA states if Cp and Cv are specific heats at constant pressure and constant volume respectively. It is observed that Cp – Cv = a for hydrogen gas and Cp – Cv = b for nitrogen gas. The correct relation between a and b is :

 (1) a= (1/14)b (2) a=b (3) a = 14b (4) a= 28b

#### Solution

According to the problem:
Let molar heat capacity at constant pressure = Xp
and molar heat capacity at constant volume = Xv
Xp – Xv = R
MCp – MCv = R
Cp – Cv = R/M
For hydrogen; a = R/2
For N2 ; b = R/28
a/b = 14
a = 14b

#### Answer to IIT JEE Main 2017 Physics Question 10 SetA

Correct answer to given problem No 10 of IIT JEE Main Physics paper of 2017 SetA is option number (3) that the correct relation between a and b is a = 14b.

## IIT JEE Main 2017 Physics Question 9 SetA:

IIT JEE Main 2017 Physics Question 9 SetA:- An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and α is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by:

 (1) P/(3αK) (2) P/(αK) (3) (3α)/(PK) (4) 3αPK

#### Solution

According to the problem:
K = ΔP/(-ΔV/V)
ΔV/V = P/K
therefore, V = V0 (1 + γΔt)
ΔV/V0 = γΔt
therefore,P/K = γΔt
=> Δt = P/γK
=> Δt = P/3αK

#### Answer to IIT JEE Main 2017 Physics Question 9 SetA

Correct answer to given problem No 9 of IIT JEE Main Physics paper of 2017 SetA is option number (1) that temperature should be raised by Δt = P/3αK.

## IIT JEE Main 2017 Physics Question 8 SetA

IIT JEE Main 2017 Physics Question 8 SetA:

A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by :
(Given : room temperature = 30°C, specific heat of copper = 0.1 cal/gm°C):

 (1) 800 °C (2) 885 °C (3) 1250 °C (4) 825 °C

#### Solution

According to the problem:
100 × 0.1 × (t – 75) = 100 × 0.1 × 45 + 170 × 1 × 45
=> 10t – 750 = 450 + 7650
=> 10t = 1200 + 7650
=> 10t = 8850
=> t =  885°C

#### Answer to IIT JEE Main 2017 Physics Question 8 SetA

Correct answer to given problem No 8 of IIT JEE Main Physics paper of 2017 SetA is option number (2) that is the initial temperature of copper ball is found to be 885 °C.

## IIT JEE Main 2017 Physics Question 7 SetA

IIT JEE Main 2017 Physics Question 7 SetA:

The variation of acceleration due to gravity g with distance d from center of the earth is best represented by (R = Earth’s radius) :

 (1) (2) (3) (4)

#### Solution

The problem can be divided into three parts as follows:
(a) Inside earth i.e. below earth’s surface: d<R,
for which acceleration due to gravity is given by
g = Gm.d/R2

(b) On Earth’s surface: d = R,
for which acceleration due to gravity is given by
gs = Gm/R2

(c) Above earth i.e. above earth’s surface: d > R,
for which acceleration due to gravity is given by
g = Gm/d2

Hence option number 4 seems to be nearest to the given case.

#### Answer to IIT JEE Main 2017 Physics Question 7 SetA

Correct answer to given problem No 7 of IIT JEE Main Physics paper of 2017 SetA is option number (4).

## IIT JEE Main 2017 Physics Question 6 SetA

IIT JEE Main 2017 Physics Question 6 SetA:

A slender uniform rod of mass M and length l is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle θ with the vertical is:

 (1) (3g/2l) sin θ (2) (2g/3l) sin θ (3) (3g/2l) cos θ (4) (2g/3l) cos θ

#### Solution

Torque at angle θ
τ = Mg sin θ (l/2)
τ =
= Mg sin θ (l/2)

as I = (Ml2)/3
[(Ml2)/3]α = Mg sin θ (l/2)
/3 = g sin (θ/2)
=> α = (3g sinθ)/2l

#### Answer to IIT JEE Main 2017 Physics Question 6 SetA

Correct answer to given problem No 6 of IIT JEE Main Physics paper of 2017 SetA is option number (1) that is angular acceleration of the rod when it makes an angle θ with the vertical is (3g sinθ)/2l.

## IIT JEE Main 2017 Physics Question 5 SetA

IIT JEE Main 2017 Physics Question 5 SetA:

The moment of inertia of a uniform cylinder of length l and radius R about its perpendicular bisector is I.
What is the ratio l/R  such that the moment of inertia is minimum?:

 (1) √(3/2) (2) (√3)/2 (3) 1 (4) 3/√2

#### Solution

Given situation in problem is depicted in the image below:

the moment of inertia is given by:
I = mR2/4 + ml2/12
=> I = (m/4)[R2 + l2/3]
=> I = (m/4)[v/Πl + l2/3]

To minimise obtain dI/dl and set it equal to zero
=> dI/dl = (m/4)[-v/Πl2 + 2l/3] = 0
=> -v/Πl2 = 2l/3
=> v = 2Πl3/3
=> ΠR2l= 2Πl3/3
=> l2 / R2= 3/2
=> l/R = √(3/2)

#### Answer to IIT JEE Main 2017 Physics Question 5 SetA

Correct answer to given problem No 5 of IIT JEE Main Physics paper of 2017 SetA is option number (1) that the ratio l/R  such that the moment of inertia is minimum will be √(3/2).

## IIT JEE Main 2017 Physics Question 4 SetA

IIT JEE Main 2017 Physics Question 4 SetA:

A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 second will be:

 (1) 4.5 J (2) 22 J (3) 9 J (4) 18 J

#### Solution

Given in problem that:
the mass of the particle is m = 1 kg,
initially the particle is at rest i.e. u =0
and let the final velocity achieved by particle be v after a time t.
then as we know that, Force = mass x acceleration, we have
=> ∫v0 dv = ∫ 6t dt
=> v = 6[t2/2]10
=> v = 3 ms-1
Thus Work done, W = change in Kinetic Energy = (1/2)* m * v * v
=> W = 0.5 * 1 * 3 * 3
=> W = 4.5 J

#### Answer to IIT JEE Main 2017 Physics Question 4 SetA

Correct answer to given problem No 4 of IIT JEE Main Physics paper of 2017 SetA is option number (1) that is the work done by the force will be 4.5 J.

## IIT JEE Main 2017 Physics Question 3 SetA

IIT JEE Main 2017 Physics Question 3 SetA:

A body of mass m = 10–2 kg is moving in a medium and experiences a frictional force F = –kv2. Its initial speed is v0 = 10 ms–1 . If, after 10 s, its energy is 1/8mv20, the value of k will be:

 (1) 10–3 kg m–1 (2) 10–3 kg s–1 (3) 10–4 kg m–1 (4) 10–1 kg m–1 s–1

#### Solution

Given in problem that the body of mass m is experiencing a frictional force of F = –kv2. Its initial speed is v0 = 10 ms–1. If, after 10 s, its energy is 1/8mv20, the value of k can be obtained by following method:
kf/ki = 1/8mv20/1/2mv20,
=> kf/ki = 1/4
=> vf/vi = 1/2
=> vf = v0/2
=> -kv2 = m dv/dt
=> ∫v0v0/2 dv/v2 = ∫0t0-k dt/m
=> 1/v0 – 2//v0 = -kt0/m
=> 1/v0 = -kt0/m
=> k = m/(v0t0)
=> k = 10-2/(10 x 10)
=> k = 10-4 kg m-1

#### Answer to IIT JEE Main 2017 Physics Question 3 SetA

Correct answer to given problem No 3 of IIT JEE Main Physics paper of 2017 SetA is option number (3) that is the value of k is 10–4 kg m–1.

## IIT JEE Main 2017 Physics Question 2 Set A

IIT JEE Main 2017 Physics Question 2 Set A:

A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time?:

 (1) (2) (3) (4)

#### Solution

Given in problem that the ball is thrown up, let us say with an initial velocity, say v. As the ball travels up, the acceleration due to gravity, opposes the motion of the ball. The magnitude of acceleration due to gravity, can be safely assumed to remain constant. Thus, results in reduction of velocity of the ball to zero, by the time it reaches top of climb. Thereafter, the velocity starts increasing, however in opposite direction, that is towards the earth. Hence, the most suitable graph representing the change in velocity is shown in option number 3.

#### Answer to IIT JEE Main 2017 Physics Question 2 Set A

Correct answer to given problem No 2 of IIT JEE Main Physics paper of 2017 Set A is option number (3) that is the velocity of the ball reduces from an initial value to a final negative value through zero.