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Test: BITSAT Past Year Paper- 2013 - JEE MCQ


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30 Questions MCQ Test - Test: BITSAT Past Year Paper- 2013

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Test: BITSAT Past Year Paper- 2013 - Question 1

The velocity and acceleration vectors of a particle undergoing circular motion are and respectively at an in stant of time. The radius of the circle is –

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 1

It can be observed th at component of acceleration perpendicular to velocity is a = 4 m/s2

Test: BITSAT Past Year Paper- 2013 - Question 2

A man runs at a speed of 4 m/s to overtake a standing bus. When he is 6 m behind the door at t = 0, the bus moves forward and continuous with a constant acceleration of 1.2 m/s2. The man reaches the door in time t. Then,

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 2

Let us draw the figure for given situation,


⇒ 4 t = 6 + 0.6 t2

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Test: BITSAT Past Year Paper- 2013 - Question 3

Wave pulse can travel along a tense string like a violin spring. A series of experiments showed that the wave velocity V of a pulse depends on the following quantities, the tension T of the string, the cross-section area A of the string and then as per unit volume ρ of the string. Obtain an expression for V in terms of the T, A and r using dimensional analysis.

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 3

Let  V = kTa Abρc,
k = dimensional constant
Writing dimension on both we side
[LT–1] = [MLT–2]a [L2]b [ML–3]c              = [Ma+cLa+2b–3cT–2a]Comparing power on both sides we have a + c = 0,   a + 2b – 3c = 1,    –2a = –1

Test: BITSAT Past Year Paper- 2013 - Question 4

A body is projected, making an acute angle with the horizontal. If angle between velocity and acceleration is θ, then

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 4

Here velocity is acting upwards wh en projectile is going upwards and acceleration is downwards. The angle θ between and ar is more than 0º and less than 180º.

Test: BITSAT Past Year Paper- 2013 - Question 5

The minimum velocity (in ms–1) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 5

The condition to avoid skidding,

Test: BITSAT Past Year Paper- 2013 - Question 6

A bob is hanging over a pulley inside a car through, a string. The second end of the string is in the hand of a person standing in the car. The car is moving with constant acceleration 'a' directed horizontally as shown in figure. Other end of the string is pulled with constant acceleration ‘a’ vertically. The tension in the string is equal to –

Test: BITSAT Past Year Paper- 2013 - Question 7

A block of mass m is placed on a smooth inclined wedge ABC of inclination q as shown in the figure. The wedge is given an acceleration ‘a’ towards the right. The relation between a and θ for the block to remain stationary on the wedge is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 7

Let the mass of block is m. It will remains stationary if forces acting on it are in equilibrium. i.e., ma cos θ = mg sin θ
⇒ a = g tan θ

Here ma = Pseudo force on block, mg = weight.

Test: BITSAT Past Year Paper- 2013 - Question 8

A 3.628 kg freight car moving along a horizontal rail road spur track at 7.2 km/hour strikes a bumper whose coil springs experiences a maximum compression of 30 cm in stopping the car. The elastic potential energy of the springs at the instant when they are compressed 15 cm is

Test: BITSAT Past Year Paper- 2013 - Question 9

A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses 0.36 kg and 0.72 kg. Taking g = 10 m/s2, find the work done (in joules) by the string on the block of mass 0.36 kg during the first second after the system is released from rest.

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 9

Given m = 0.36 kg, M = 0.72 kg.
The figure shows the forces on m and M.
When the system is released, let the acceleration be a. Then

T – mg = ma
Mg – T = Ma
∴ 
and T =  4 mg/3
For block m :
u = 0, a = g/3, t = 1, s = ?

∴ Work done by the string on m is

Test: BITSAT Past Year Paper- 2013 - Question 10

Two rings of radius R and nR made of same material have the ratio of moment of inertia about an axis passing through centre is 1 : 8. The value of n is 

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 10

Ratio of moment of inertia of the rings

(λ = linear density of wire = constant)

∴ n3 = 8 ⇒ n = 2

Test: BITSAT Past Year Paper- 2013 - Question 11

A particle of mass ‘m’ is projected with a velocity v making an angle of 30° with the horizontal. The magnitude  of angular momentum of the projectile about the point of projection when the particle is at its maximum  height ‘h’ is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 11

Angular momentum L0 = pr ⊥
(∴ linear momentum p = mv cos θ and r = H)
⇒ L0 = mv cos θH


Test: BITSAT Past Year Paper- 2013 - Question 12

There is a shell of mass M and density of the shell is uniform. The work done to take a point mass from point A to B is  [AB = r]

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 12

Gravitaional field inside the shell is zero, so no work required.

Test: BITSAT Past Year Paper- 2013 - Question 13

A disc is performing pure rolling on a smooth stationary surface with constant angular velocity as shown in figure. At any instant, for the lower most point of the disc –

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 13

As the disc is in combined rotation and translation, each point has a tangential velocity and a linear velocity in the forward direction. From figure

vnet (for lowest point) = v – Rω = v – v = 0 and acceleration

(since linear speed is constant)

Test: BITSAT Past Year Paper- 2013 - Question 14

A cube is subjected to a uniform volume compression. If the side of the cube decreases by 2% the bulk strain is

Test: BITSAT Past Year Paper- 2013 - Question 15

A ball whose density is 0.4 × 103 kg/m3 falls into water from a height of 9 cm. To what depth does the ball sink ?

Test: BITSAT Past Year Paper- 2013 - Question 16

Figure shows a copper rod joined to a steel rod.
The rods have equal length and equal crosssectional area. The free end of the copper rod is kept at 0ºC and that of steel rod is kept at 100ºC.
Find the temperature of the junction of the rod.
Conductivity of copper = 390 W/mºC.
Conductivity of steel = 46 W/m ºC

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 16

Heat current in first rod (copper)

Here q is ltemperature of the junction and A & l are area and length of copper rod.
Heat current in second rod (steel)

In series clombination, heat current remains same. So,

– 390 θ = = 46 θ – 4600 436 θ = 4600  ⇒  θ = 10.6ºC

Test: BITSAT Past Year Paper- 2013 - Question 17

If the radius of a star is R and it acts as a black body, what would be the temperature of the star, in which the rate of energy production is Q ?

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 17

Stefan’s law for black body radiation
Q = σe AT4

Here e = 1
A = 4πR2

Test: BITSAT Past Year Paper- 2013 - Question 18

A thermodynamical system is changed from state (P1, V1) to (P2, V2) by two different process, the quantity which will remain same will be

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 18

For all process
ΔU = ΔQ – ΔW
does not change as it depends on initial final states.

Test: BITSAT Past Year Paper- 2013 - Question 19

A Car not’s heat engine works between the temperatures 427°C and 27°C. What amount of heat should it consume per second to deliver mechanical work at the rate of 1.0 kW?

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 19

The efficiency of the heat engine is

But 

Thus, the engine would require 417 cal of heat per second, to deliver the requisite amount of work.

Test: BITSAT Past Year Paper- 2013 - Question 20

A vessel containing 1 more of O2 gas (molar mass 32) at temperature T. The pressure of the gas is p.An identical vessel containing one mole of he gas (molar mass 4) at temperature 2T has a pressure of

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 20

Applying gas equation, pV = nRT We can write, p1V = n1RT1 and p2V = n2RT2

Test: BITSAT Past Year Paper- 2013 - Question 21

The temperature of an ideal gas is increased from 27°C to 127°C, then percentage increase in vrms is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 21

We know, 
⇒ % increase in

Test: BITSAT Past Year Paper- 2013 - Question 22

Two gases occupy two containers A and B the gas in A, of volume 0.10m3, exerts a pressure of 1.40 MPa and that in B of volume 0.15m3 exerts a pressure 0.7 MPa. The two containers are united by a tube of negligible volume and the gases are allowed to intermingle. Then if the temperature remains constant, the final pressure in the container will be (in MPa)

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 22

We know that
PAVA = nA RT, PBVB = nBRT and Pf (VA + VB) = (nA + nB) RT
Pf (VA + VB) = PAVA + PBVB

Test: BITSAT Past Year Paper- 2013 - Question 23

An instan tan eous displacemen t of a simple harmonic oscillator is x = A cos (ωt + p/4). Its speed will be maximum at time

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 23

Velocity, 
Velocity will be maximum, when wt + ω/4 = π/2 or ωt = π/2 – π/4 = π/4
or t = π/4ω

Test: BITSAT Past Year Paper- 2013 - Question 24

Two waves of wavelengths 99 cm and 100 cm both travelling with velocity 396 m/s are made to interfere. The number of beats produced by them per second is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 24

Velocity of wave v = nλ
where n = frequency of wave 

no. of beats = n1 – n2 = 4

Test: BITSAT Past Year Paper- 2013 - Question 25

If equation of transverse wave is y = x0 cos Maximum velocity of particle istwice of wave velocity, if l is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 25


Test: BITSAT Past Year Paper- 2013 - Question 26

Three equal charges (q) are placed at corners of an equilateral triangle of side a. The force on any charge is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 26

Test: BITSAT Past Year Paper- 2013 - Question 27

Two identical capacitors, have the same capacitance C. One of them is charged to potential V1 and the other to V2. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is –

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 27

Initial energy of combined system

Final common potential, 
Final energy of system,

Hence loss of energy = U1 – U2

Test: BITSAT Past Year Paper- 2013 - Question 28

What should be the characteristic of fuse wire?

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 28

Fuse wire should be such that it melts immediatley when strong current flows through the circuit. The same is possible if its melting point is low and resistivity is high.

Test: BITSAT Past Year Paper- 2013 - Question 29

In the circuit shown in figure potential difference between points A and B is 16 V. the current passing through 2Ω resistance will be​

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 29

∴ 4i1 + 2(i1 + i2) – 3 + 4i1 = 16V ...(i) Using Kirchhoff’s second law in the closed loop we have
9 – i2 – 2(i1 + i2) = 0 …(ii)
Solving equations (i) and (ii), we get
i1 = 1.5 A   and i2 = 2 A
∴ current through 2W resistor = 2 + 1.5 = 3.5 A.

Test: BITSAT Past Year Paper- 2013 - Question 30

Two parallel conductors carry current in opposite directions as shown in figure. One conductor carries a current of 10.0 A. Point C is a distance d/2 to the right of the 10.0 A current. If d = 18 cmand I is adjusted so that the magnetic field at C is zero, the value of the current I is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 30

The magnetic field at C due to first conductor is (since, point C is separated by from 1st conductor). The direction of field is perpendicular to the plane of paper and directed outwards.
The magnetic field at C due to second conductor is (since, point C is separated by d/2 from 2nd conductor)
The direction of field is perpendicular to the plane of paper and directed inwards.
Since, direction of B1 and B2 at point C is in opposite direction and the magnetic field at C is zero, therefore,
 B1 = B2

On solving I = 30.0 A

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