IIT JEE Main 2016 Physics Question 30

IIT JEE Main 2016 Physics Question 30

A pendulum clock loses 12 s a day if the temperature is 40ºC and gains 4s a day if the temperature is 20ºC. The temperature at which the clock will show correct time, and the coefficient of linear expansion (α) of the metal of the pendulum shaft are respectively:

(1) 30ºC; α = 1.85 × 10-3/ºC (2) 55ºC; α = 1.85 × 10-2/ºC (3) 25ºC; α = 1.85 × 10-5/ºC (4) 60ºC; α = 1.85 × 10-4/ºC

Solution

(ΔT/T) =(ΔL/L)
(ΔT/T) = (αΔT/2)
=> (ΔT/T) = (αΔT1/2)
=> (1/2)(αΔT1)x24x3600
=> 12 = (αΔT1) x 12 x 3600
=> 4 = (αΔT2) x 12 x 3600
=> 3 = (ΔT1/ΔT2) = (40 – T)/(T-20)
=> 3T – 60 = 40 – T
=> T = 25ºC
α = 1.85 × 10-5/ºC

Answer

Correct answer to given problem No 30 of IIT JEE Main Physics paper of 2016 is option number (3).

IIT JEE Main 2016 Physics Question 29

IIT JEE Main 2016 Physics Question 29

In an experiment for determination of refractive index of glass of a prism by i – δ, plot, it was found that a ray incident at angle 35º, suffers a deviation of 40º and that it emerges at angle 79º. In that case which of the following is closest to the maximum possible value of the refractive index?

(1) 1.7 (2) 1.8 (3) 1.5 (4) 1.6

Solution

∠i = 35º
δ = 40º
e = 79º
i = 35º
e = 79º
δ = 40º
δ + 4 = i + e
=> A = 74º
with above set of data, as refractive index increases emerging ray will have tendency to suffer TIR.
For i = 35º
μ = sin 35º/ sin r1
=> μ = sin 90º / sin (74º – r1)
Using reversibility principle
For i = 79º
μ = sin 79º/ sin r2 = sin 90º / sin (74º – r2)
From 2nd case r2 = 37º
=> μ =1.63

Answer

Correct answer to given problem No 29 of IIT JEE Main Physics paper of 2016 is option number (4).

IIT JEE Main 2016 Physics Question 28

IIT JEE Main 2016 Physics Question 28

An observer looks at a distance tree of height 10m with a telescope of magnifying power of 20. To the observer the tree appears:

(1) 20 times taller (2) 20 times nearer (3) 10 times taller (4) 10 times nearer

Solution

By definition,  for telescopes, magnifying power = 20 means that the object appears 20 times nearer.

Answer

Correct answer to given problem No 28 of IIT JEE Main Physics paper of 2016 is option number (2).

IIT JEE Main 2016 Physics Question 27

IIT JEE Main 2016 Physics Question No 27

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 PVn constant, then n is given by (Here Cp and Cv are molar specific heat at constant pressure and constant volume, respectively):

(1) n= (Cp – C)/(C- Cv) (2)n= (C- Cv)/(C – Cp) (3) n= Cp / Cv (4) (C- Cp)/(C – Cv)

Solution

PVn = constant
Then Cprocess for the above polytropic process is known
Cprocess = C = Cv + R/(1-n)
C- Cv = R/(1-n)
(1-n) = R / (C- Cv)
So after solving we get,
n = (C- Cv – R)/(C -Cv)
=> n = (C- Cp)/(C – Cv)

Answer

Correct answer to given problem No 27 of IIT JEE Main Physics paper of 2016 is option number (4) that is (C- Cp)/(C – Cv).

IIT JEE Main 2016 Physics Question 26

IIT JEE Main 2016 Physics Question No 26

A uniform string of length 20 m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is: (take 2 g 10 ms-2 ).

(1) 2√2s (2) √2s (3) 2Π√2s (4) 2s

Solution

ν = √[(μxg)/μ]
= √(xg)
dx/dt = √(xg)

0-1 [dx/√(x)] = t√(g)
2x1/2 = t√(g)
[2/√(g)]√(20) = t
2√2 = t

Answer

Correct answer to given problem No 26 of IIT JEE Main Physics paper of 2016 is option number (1) that is 2√2.

Save

Save

IIT JEE Main 2016 Physics Question 25

IIT JEE Main 2016 Physics Question No 25

The box of a pin hole camera, of length L, has a hole of radius a. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength λ the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say bmin) when:

(1) a=√(λL) and bmin=√(4λL) (2) a=λ2/L and bmin=√(4λL) (3) a=λ2/L and bmin=(2λ2/L) (4) a=√(λL) and bmin=(2λ2/L)

Solution

IIT JEE Main 2016 Physics Question 25

b sin θ = nλ
(b-a)/L = tan θ ≈ sin θ
a(b-a)/L = λ
ab – a2 = λL
a2 -ab + λL = 0
b2 >= 4λL
bmin = √(4λL)
a(√(4λL)) – a2 = λL
a = λ2/L

Answer

Correct answer to given problem No 25 of IIT JEE Main Physics paper of 2016 is option number (2).

IIT JEE Main 2016 Physics Question 24

IIT JEE Main 2016 Physics Question No 24

Hysteresis loops for two magnetic material A and B are given below:
IIT JEE Main Physics Question 24
These materials are used to make magnets for electric generators, transformer core and electromagnet core. Then it is proper to use:
(1) A for transformers and B for electric generators
(2) B for electromagnets and transformers
(3) A for electric generators and transformers
(4) A for electromagnets and B for electric generators

Solution

Option no 2 as Hysteresis loss for electromagnets and transformers are required to be less, so that, it can magnetize or demagnetize easily.

Answer

Correct answer to given problem No 24 of IIT JEE Main Physics paper of 2016 is option number (2).

IIT JEE Main 2016 Physics Question 23

IIT JEE Main 2016 Physics Question No 23

A roller is made by joining together two cones at their vertices O. It is kept on two rails AB and CD which are placed asymmetrically (see figure), with its axis perpendicular to CD and its centre O at the centre of line joining AB and CD (see figure). It is given a light push so that it starts
rolling with its centre O moving parallel to CD in the direction shown. As it moves, the roller will tend to:

(1) go straight (2) turn left and right alternately (3) turn left (4) turn right

Solution

IIT JEE Main 2016 Physics Question 23

Initially C.M. will move straight, hence its distances from left rail will decrease.
Normal reaction at left contact N1 will increase. Horizontal component of left N1x will also
increase and continue to increase. Hence the roller will tend to move towards left direction.

Answer

Correct answer to given problem No 23 of IIT JEE Main Physics paper of 2016 is option number (3).

IIT JEE Main 2016 Physics Question 22

IIT JEE Main 2016 Physics Question No 22

A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a time sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with
the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?

(1) 0.70 mm (2) 0.50 mm (3) 0.75 mm (4) 0.80 mm

Solution

We know that,
Least Count (LC) = pitch/total divisions
Here, LC = 0.5/50
=> LC = 0.01 mm
Here the error is negative and is given by
error = 5 × 0.01 = 0.05
Reading = 0.5 + 25 × 0.01 + 0.05
= 0.5 + 0.25 + 0.05
= 0.80 mm

Answer

Correct answer to given problem No 22 of IIT JEE Main Physics paper of 2016 is option number (4) that is 0.80 mm.