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Test: BITSAT Past Year Paper- 2010 - JEE MCQ


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30 Questions MCQ Test - Test: BITSAT Past Year Paper- 2010

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Test: BITSAT Past Year Paper- 2010 - Question 1

If P represents radiation pressure, c represents speed of light and Q represents radiation energy striking a unit area per second, the non-zero integers x, y and z such that PxQycz is dimensionless, are.

Test: BITSAT Past Year Paper- 2010 - Question 2

The position x of a particle varies with time (t) as x = A t2 – B t3. The acceleration at time t of the particle will be equal to zero. What is the value of t?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 2

Given that x = A t2 – B t3
∴ velocity dx/dy = 2A t - 3 Bt 2
and  acceleration =   = 2 A - 6Bt
For acceleration to be zero 2A – 6Bt = 0.

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Test: BITSAT Past Year Paper- 2010 - Question 3

Two projectiles A and B are thrown with the same speed but angles are 40º and 50º with the horizontal. Then

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 3

lesser is the value of θ, lesser is sinq and hence lesser will be the time taken. Hence A will fall earlier.

Test: BITSAT Past Year Paper- 2010 - Question 4

A body is travelling in a circle at a constant speed. It

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 4

Body moves with constant speed it means that tangential acceleration aT = 0 & only centripetal acceleration aC exists whose direction is always towards the centre or inward (along the radius of the circle).

Test: BITSAT Past Year Paper- 2010 - Question 5

Two blocks are connected over a massless pulley as shown in fig. The mass of block A is 10 kg and the coefficient of kinetic friction is 0.2. Block A slides down the incline at constant speed. The mass of block B in kg is:

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 5

Considering the equilibrium of A, we get 10a = 10g sin 30º – T – mN
where N = 10g cos 30°


but a = 0, T = mBg

⇒ mB = 3.268 ≈ 3.3 kg

Test: BITSAT Past Year Paper- 2010 - Question 6

A spring is compressed between two toy carts of mass m1 and m2. When the toy carts are released, the springs exert equal and opposite average forces for the same time on each toy cart. If v1 and v2 are the velocities of the toy carts and there is no friction between the toy carts and the ground, then :

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 6

Applyin g law of conservation of lin ear momentum m1v1 + m2v2 = 0,

 

Test: BITSAT Past Year Paper- 2010 - Question 7

The potential energy for a force field is given by U (x, y) = cos (x + y). The force acting on a particle at position given by coordinates (0, π/4) is –

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 7

Test: BITSAT Past Year Paper- 2010 - Question 8

A long string is stretched by 2 cm and the potential energy is V. If the spring is stretched by 10 cm, its potential energy will be

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 8

Test: BITSAT Past Year Paper- 2010 - Question 9

The ratio of the accelerations for a solid sphere (mass ‘m’ and radius ‘R’) rolling down an incline of angle ‘θ’ without slipping and slipping down the incline without rolling is:

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 9

For solid sphere rolling without slipping on inclined plane, acceleration

For solid sphere slipping on inclined plane without rolling, acceleration
a2 = g sin θ
Therefore required ratio = 

Test: BITSAT Past Year Paper- 2010 - Question 10

A system consists of three particles, each of mass m and located at (1, 1), (2, 2) and (3, 3). The co-ordinates of the centre of mass are

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 10

The coordinates of C.M of three particle are

so coordinates of C.M. of three particle are (2, 2) V = 300 m/s

Test: BITSAT Past Year Paper- 2010 - Question 11

Suppose the gravitational force varies inversely as the nth power of distance. Then the time period of a planet in circular orbit of radius ‘R’ around the sun will be proportional to

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 11

F = KR- n = MRω2 ⇒ ω2 = KR- (n +1)

[where K' = K1/2, a constant]

Test: BITSAT Past Year Paper- 2010 - Question 12

Two planets A and B have the same material density. If the radius of A is twice that of B, then the ratio of the escape velocity vA/vB is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 12


as ve ∝ R for same density, 

Test: BITSAT Past Year Paper- 2010 - Question 13

The upper end of a wire of diameter 12mm and length 1m is clamped and its other end is twisted through an angle of 30°. The angle of shear is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 13

Test: BITSAT Past Year Paper- 2010 - Question 14

A spherical ball is dropped in a long column of a viscous liquid. The speed (v) of the ball as a function of time (t) may be best represented by

Test: BITSAT Past Year Paper- 2010 - Question 15

Two mercury drops (each of radius r) merge to form a bigger drop. The surface energy of the bigger drop, if T is the surface tension, is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 15


Surface energy of bigger drop,
E = 4πR2T = 4 x 22/3 πr2T = 28/3 πr2T

Test: BITSAT Past Year Paper- 2010 - Question 16

Two circular plates of radius 5 cm each, have a 0.01 mm thick water film between them. Then what will be the force required to separate these plate (S.T. of water = 73 dyne/cm) ?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 16


    = 36.5 π ≈ 115 newton

Test: BITSAT Past Year Paper- 2010 - Question 17

One kilogram of ice at 0°C is mixed with one kilogram of water at 80°C. The final temperature of the mixture is (Take specific heat of water = 4200 kJ/kg-°C, Latent heat of ice = 336 kJ/kg)

Test: BITSAT Past Year Paper- 2010 - Question 18

In the equation PVγ = constant, the value of γ is unity. Then the  process is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 18

PV = constant represents isothermal process.

Test: BITSAT Past Year Paper- 2010 - Question 19

An ideal refrigerator has a freezer at a temperature of 13ºC. The coefficient of performance of the engine is 5. The temperature of the air (to which heat is rejected) is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 19

T2 = 273 – 13 = 260,

or T1 – 260 = 52; T1 = 312 K,
T2 = 312 – 273 = 39°C

Test: BITSAT Past Year Paper- 2010 - Question 20

3 moles of an ideal gas at a temperature of 27°C are mixed with 2 moles of an ideal gas at a temperature 227°C, determine the equilibrium temperature of the mixture, assuming no loss of energy.

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 20

Energy possessed by the ideal gas at 27°C is

Energy possessed by the ideal gas at 227°C is

If T be the equilibrium temperature, of the mixture, then its energy will be

Since, energy remains conserved,
Em = E+ E2

or T = 380 K  or 107°C

Test: BITSAT Past Year Paper- 2010 - Question 21

A simple pendulum has time period 't'. Its time period in a lift which is moving upwards with acceleration 3 ms–2 is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 21

Test: BITSAT Past Year Paper- 2010 - Question 22

A wave y = a sin (ωt – kx) on a string meets with another wave producing a node at x = 0. Then the equation of the unknown wave is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 22

Equation of a wave
y1 = a sin (ωt – kx)         ....(i)
Let equations of another wave may be,
y2 = a sin (ωt + kx)         ....(ii)
y3 = –a sin (ωt + kx)      ....(iii)
If Eq. (i) propagate with Eq. (ii), we get y = 2a cos kx sin ωt
If Eq. (i), propagate with Eq. (iii), we get y = –2a sin kx cos ωt
At x = 0, y = 0, wave produce node
So, Eq.(iii) is the equation of unknown wave

Test: BITSAT Past Year Paper- 2010 - Question 23

A source of sound produces waves of wavelength 60 cm when it is stationary. If the speed of sound in air is 320 m s-1 and source moves with speed 20 m s-1, the wavelength of sound in the forward direction will be nearest to

Test: BITSAT Past Year Paper- 2010 - Question 24

A charge +q is at a distance L/2 above a square of side L. Then what is the flux linked with the surface?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 24


The given square of side L may be considered as one of the faces of a cube with edge L. Then given charge q will be considered to be placed at the centre of the cube. Then according to Gauss's theorem, the magnitude of the electric flux through the faces (six) of the cube is given by
φ = q/ε0
Hence, electric flux through one face of the cube for the given square will be

Test: BITSAT Past Year Paper- 2010 - Question 25

Two metallic spheres of radii 1 cm and 3 cm are given charges of -1 × 10-2 C and 5 × 10-2 C, respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 25

Test: BITSAT Past Year Paper- 2010 - Question 26

In a region, the potential is represented by V(x, y, z) = 6x - 8xy - 8y + 6yz, where V is in volts and x, y, z are in metres. The electric force experienced by a charge of 2 coulomb situated at point (1, 1, 1) is :

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 26

Test: BITSAT Past Year Paper- 2010 - Question 27

The power dissipated in the circuit shown in the figure is 30 Watts. The value of R is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 27

     ...(i)

Substituting the values in equation (i)

Test: BITSAT Past Year Paper- 2010 - Question 28

Which of the following quantities do not change when a resistor connected to a battery is heated due to the current?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 28

Only number of free electrons is constant, other factors are temperature dependent.

Test: BITSAT Past Year Paper- 2010 - Question 29

The magnetic field at the origin due to the current flowing in the wire is –

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 29

Test: BITSAT Past Year Paper- 2010 - Question 30

The back emf induced in a coil, when current changes from 1 ampere to zero in one milli-second, is 4 volts, the self inductance of the coil is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 30

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