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Test: BITSAT Past Year Paper- 2014 - JEE MCQ


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30 Questions MCQ Test Additional Study Material for JEE - Test: BITSAT Past Year Paper- 2014

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Test: BITSAT Past Year Paper- 2014 - Question 1

A rifle man, who together with his rifle has a mass of 100 kg, stands on a smooth surface and fires 10 shots horizontally. Each bullet has a mass 10 g and a muzzle velocity of  800 ms–1. The velocity which the rifle man attains after  firing 10 shots is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 1

According to law of conservation of
momentum,
100v = -1/1000 x 10 x 800
i.e., v = 0.8 ms–1.

Test: BITSAT Past Year Paper- 2014 - Question 2

A train accelerating uniformly from rest attains a maximum speed of 40 ms–1 in 20 s. It travels at the speed for 20 s and is brought to rest with uniform retardation in further 40 s. What is the average velocity during the period ?

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 2

(i) v = u + at1
40 = 0 + a × 20
a = 2 m/s2
v2 – u2 = 2as
402 – 0 = 2 × 2 s1
s1 = 400 m
(ii) s2 = v × t2 = 40 × 20 = 800 m
(iii) v = u + at
0 = 40 + a × 40
a = –1 m/s2
02 – 402 = 2(–1)s3
s3 = 800 m
Total distance travelled = s1 + s2 + s3
= 400 + 800 + 800 = 2000 m
Total time taken = 20 + 20 + 40 = 80 s
Average velocity 2000/80 = 25m/s

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Test: BITSAT Past Year Paper- 2014 - Question 3

A projectile is fired with a velocity u making an angle q with the horizontal. What is the magnitude of change in velocity when it is at the highest point –

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 3

Initially u = cosθ iˆ + usinθ ˆj .
At highest point v = u cosθ iˆ
∴ difference is u sin q.

Test: BITSAT Past Year Paper- 2014 - Question 4

For the equation F = Aavbdc, where F is the force, A is the area, v is the velocity and d is the density, the values of a, b and c are respectively

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 4

[MLT–2] = [L2a] × [LbT–b][McL–3c]
= [McL2a + b –3cT–b]
Comparing powers of M, L and T, on both
sides, we get
c = 1, 2a + b –3c = 1, –b = –2 or b = 2
Also, 2a + 2 – 3(1) =1 Þ 2a = 2 or a = 1
∴ This is 1, 2, 1

Test: BITSAT Past Year Paper- 2014 - Question 5

A person with his hand in his pocket is skating on ice at the rate of 10m/s and describes a circle of radius 50 m. What is his inclination to vertical: (g = 10 m/sec2)

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 5

Since surface (ice) is frictionless, so the
centripetal force required for skating will be
provided by inclination of boy with the
vertical and that angle is given as
tan θ = v2/rg
= where v is speed of skating &
r is radius of circle in which he moves.

Test: BITSAT Past Year Paper- 2014 - Question 6

A small block of mass m is kept on a rough inclined surface of inclination q fixed in a elevator. The elevator goes up with a uniform velocity v and the block does not slide on the wedge. The work done by the force of friction on the block in time t will be :

Test: BITSAT Past Year Paper- 2014 - Question 7

An equilateral prism of mass m rests on a rough horizontal surface with coefficient of friction µ.

A horizontal force F is applied on the prism as shown in the figure.
If the coefficient of  friction is sufficiently high so that the prism does not slide before toppling, then the minimum force required to topple the prism is –

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 7

The tendency of rotation will be about the point C.

For minimum force, the torque of F about C has to be equal to the torque of mg about C.
∴ 

Test: BITSAT Past Year Paper- 2014 - Question 8

A spherically symmetric gravitational system of particles has a mass density  where r0 is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed V as a function of distance r (0 < r < ∞) from the centre of the system is represented by

Test: BITSAT Past Year Paper- 2014 - Question 9

The load versus elongation graph for four wires is shown. The thinnest wire is

Test: BITSAT Past Year Paper- 2014 - Question 10

Th e wor k done in blowin g a soap bubble of surface tension 0.06 × Nm–1 from 2 cm radius to 5 cm radius is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 10


= 0.003168 J

Test: BITSAT Past Year Paper- 2014 - Question 11

The wavelength of radiation emitted by a body depends upon

Test: BITSAT Past Year Paper- 2014 - Question 12

One mole of O2 gas having a volume equal to 22.4 Litres at 0°C and 1 atmospheric pressure in compressed isothermally so that its volume reduces to 11.2 litres. The work done in this process is

Test: BITSAT Past Year Paper- 2014 - Question 13

In a thermodynamic process, the pressure of a fixed mass of a gas is changed in such a manner that the gas releases 20 J of heat and 8 J of work is done on the gas. If the initial internal energy of the gas was 30 J, then the final internal energy will be

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 13

Given ΔQ = – 20 J, W = – 8 J
Using Ist law ΔQ = ΔU + ΔW
⇒ ΔQ = – 20 + 8 = – 12 J
Uf = – 12 + 30 = 18 J

Test: BITSAT Past Year Paper- 2014 - Question 14

In the kinetic theory of gases, which of these statements is/are true ?
(i) The pressure of a gas is proportional to the mean speed of the molecules.
(ii) The root mean square speed of the molecules is proportional to the pressure.
(iii) The rate of diffusion is proportional to the mean speed of the molecules.
(iv) The mean translational kinetic energy of a gas is proportional to its kelvin temperature.

Test: BITSAT Past Year Paper- 2014 - Question 15

Two balloons are filled one with pure he gas and other with air respectively. If the pressure and temperature of these balloons are same, then the number of molecules per unit volume is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 15

Assumingthe balloons have the same volume, as pV= nRT. If P, V and T are the same, n the number of moles present will be the same, whether it is He or air. 
Hence, number of molecules per unit volume will be same in both the balloons.

Test: BITSAT Past Year Paper- 2014 - Question 16

Two particles P and Q describe S.H.M. of same amplitude a, same frequency f along the same straight line. The maximum distance between the two particles is a√2 . The initial phase difference between the particle is –

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 16

x1 = a sin(ωt + ϕ1), x2 = a sin(ωt + ϕ2 )

⇒ 

To maximize 

⇒ 

⇒ 

⇒ 

Test: BITSAT Past Year Paper- 2014 - Question 17

A tunnel has been dug through the centre of the earth and a ball is released in it. It executes S.H.M. with time period

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 17

=  84.6 min

Test: BITSAT Past Year Paper- 2014 - Question 18

A soun d source, em itting sound of con stan t frequency, moves with a constant speed and crosses a stationary observer. The frequency (n) of sound heard by the observer  is plotted against time (t). Which of the following graphs represents the correct variation ?

Test: BITSAT Past Year Paper- 2014 - Question 19

When a string is divided into three segments of length l1, l2, and l3 the fundamental frequencies of these three segments are v1, v2 and v3 respectively. The original fundamental frequency (v) of the string is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 19

Fundamental frequency is given by

Since, P divided into l, land l3 segments 
Herel = l1 + l2 + l3
So 

Test: BITSAT Past Year Paper- 2014 - Question 20

Two point dipoles pkˆ and p/2 kˆ are located at(0, 0, 0) and (1m, 0, 2m) respectively. The resultant electric field due to the two dipoles at the point (1m, 0, 0) is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 20

The given point is at axis of 
dipole and at equatorial line of dipole so that field at given point.

Test: BITSAT Past Year Paper- 2014 - Question 21

Electric field in the region is given by  then the correct expression for the potential in the region is [assume potential at infinity is zero]

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 21

Test: BITSAT Past Year Paper- 2014 - Question 22

Three capacitors C1 = 1 µF, C2 = 2 µF and C3 = 3 µF are connected as shown in figure, then the equivalent capacitance between points A and B is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 22

Ceq = (1 + 2 + 3)μF = 6 μF

Test: BITSAT Past Year Paper- 2014 - Question 23

Two long coaxial and conducting cylinders of radius a and b are separated by a material of conductivity s and a constant potential difference V is maintained between them, by a battery. Then the current, per unit length of the cylinder flowing from one cylinder to the other is –

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 23

Now  

From (1) : 

Test: BITSAT Past Year Paper- 2014 - Question 24

A wire X is half the diameter and half the length of a wire Y of similar material. The ratio of resistance of X to that of Y is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 24

Test: BITSAT Past Year Paper- 2014 - Question 25

A narrow beam of protons and deuterons, each having the same momentum, enters a region of uniform magnetic field directed perpendicular to their direction of momentum. The ratio of the radii of the circular paths described by them is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 25

Since, the radius of circular path of a charged particle in magnetic field is 
Now, the radius of circular path of charged particle of given momentum r and magnetic field B is given by 

Test: BITSAT Past Year Paper- 2014 - Question 26

For th e cir cuit (figur e), the cur rent is to be measured. The ammeter shown is a galvanometer with a resistance RG = 60.00W converted to an ammeter by a shunt resistance rs = 0.02W. The value of the current is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 26

RG = 60.00W, shunt resistance, rs = 0.02W 
Total resistance in the circuit is RG + 3 = 63W
Hence, I = 3/63 = 0.048 A
Resistance of the galvanometer converted to an ammeter is,
Total resistance in the circuit = 0.02 + 3
= 3.02W
Hence, I = 3/3.02 = 0.99 A

Test: BITSAT Past Year Paper- 2014 - Question 27

The susceptibility of a magnetism at 300 K is 1.2 × 10–5. The temperature at which the susceptibility increases to 1.8 × 10–5 is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 27


⇒ 
⇒ 

Test: BITSAT Past Year Paper- 2014 - Question 28

A coil 10 turns and a resistance of 20W is connected in series with B.G. of resistance 30W. The coil is placed with its plane perpendicular to the direction of a uniform magnetic field of induction 10–2 T. If it is now turned through an angle of 60° about an axis in its plane. Find the charge induced in the coil. (Area of a coil = 10–2 m²)

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 28

Given : n = 10 turns, Rcoil = 20W, RG = 30W, Total resistance in the circuit = 20 + 30 = 50W.

= 1 × 10–5 C (Charge induced in a coil)

Test: BITSAT Past Year Paper- 2014 - Question 29

Voltage V and current i in AC circuit are given by V = 50 sin (50 t) volt, i = 50 sin. The power dissipated in the circuit is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 29

Given V = 50 sin (50 t) V
Maximum voltage, V0 = 50 V,

Maximum current, i0 = 50 mA = 50 × 10–3 A
Power dissipated, 

Test: BITSAT Past Year Paper- 2014 - Question 30

Resolving power of the telescope will be more, if the diameter of the objective is

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