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Test: BITSAT Past Year Paper- 2009 - JEE MCQ


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30 Questions MCQ Test - Test: BITSAT Past Year Paper- 2009

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Test: BITSAT Past Year Paper- 2009 - Question 1

Given that  and A2 + B2 = R2. The angle between  is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 1

Test: BITSAT Past Year Paper- 2009 - Question 2

In the relation :

P is pressure, Z is distance, k is Boltzmann constant and θ is the temperature. The dimensional formula of b will be​

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Test: BITSAT Past Year Paper- 2009 - Question 3

Which of the following is most accurate?

Test: BITSAT Past Year Paper- 2009 - Question 4

A projectile projected at an angle 30º from the horizontal has a range R. If the angle of projection at the same initial velocity be 60º, then the range will be

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 4

If sum of angle of projection = 90° for given speed then range for that angle of projection is same.

Test: BITSAT Past Year Paper- 2009 - Question 5

A block of mass M is pulled along a horizontal frictionless surface by a rope of mass M/2. If a force 2Mg is applied at one end of the rope, the force which the rope exerts on the block is –​

Test: BITSAT Past Year Paper- 2009 - Question 6

A chain of mass M is placed on a smooth table with 1/n of its length L hanging over the edge.The work done in pulling the hanging portion of the chain back to the surface of the table is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 6

W = change in PE of COM of hanging part

Test: BITSAT Past Year Paper- 2009 - Question 7

A particle of mass 10 kg moving eastwards with a speed 5 ms–1 collides with another particle of the same mass moving north-wards with the same speed 5 ms–1. The two particles coalesce on collision. The new particle of mass 20 kg will move in the north-east direction with velocity

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 7

Here


Test: BITSAT Past Year Paper- 2009 - Question 8

A uniform cube of side a and mass m rests on a rough horizontal table. A horizontal force F is applied normal to one of the faces at a point that is directly above the centre of the face, at a height 3a/4 above the base. The minimum value of F for which the cube begins to topple an edge is (assume that cube does not slide)

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 8

For toppling about edge xx'
At the moment of toppling the normal force pass through axis xx'.

Test: BITSAT Past Year Paper- 2009 - Question 9

The rotation of the earth having radius R about its axis speeds upto a value such that a man at latitude angle 600 feels weightless. The duration of the day in such case will be :

Test: BITSAT Past Year Paper- 2009 - Question 10

A metallic rod breaks when strain produced is 0.2%. The Young’s modulus of the material of the rod is 7 × 109 N/m2. What should be its area of cross-section to support a load of 104 N ?

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 10

Maximum possible strain = 0.2/100

Test: BITSAT Past Year Paper- 2009 - Question 11

A liquid is flowing through a non-sectional tube with its axis horizontally. If two points X and Y on the axis of tube has a sectional area 2.0 cm2 and 25 mm2 respectively then find the flow velocity at Y when the flow velocity at X is 10m/s.

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 11

According to principle of continuity

Test: BITSAT Past Year Paper- 2009 - Question 12

A body of length 1m having cross-sectional area 0.75m2 has heat flow through it at the rate of 6000 Joule/sec. Then find the temperature difference if K = 200 Jm–1K–1.

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 12

Test: BITSAT Past Year Paper- 2009 - Question 13

Which of the following combinations of properties would be most desirable for a cooking pot?

Test: BITSAT Past Year Paper- 2009 - Question 14

A particle starts moving rectilinearly at time t = 0 such that its velocity v changes with time t according to the equation v = t2 – t where t is in seconds and v is in m/s. Find the time interval for which the particle retards.

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 14

Acceleration of the particle a = 2t – 1
The particle retards when acceleration is opposite to velocity.

Now t is always positive
∴  (2t – 1) (t – 1) < 0

This is not possible
or  
2t – 1 > 0 & t – 1 < 0 ⇒  1/2 < t < 1

Test: BITSAT Past Year Paper- 2009 - Question 15

A sample of gas expands from volume V1 to V2.
The amount of work done by the gas is greatest when the expansion is

Test: BITSAT Past Year Paper- 2009 - Question 16

A cyclic process is shown in the p-T diagram. Which of the curves show the same process on a P-V diagram ?

Test: BITSAT Past Year Paper- 2009 - Question 17

Which one the following graphs represents the behaviour of an ideal gas

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 17

For an ideal gas PV = constant i.e., PV does not vary with V.

Test: BITSAT Past Year Paper- 2009 - Question 18

In case of a forced vibration, the resonance wave becomes very sharp when the

Test: BITSAT Past Year Paper- 2009 - Question 19

A pendulum bob carries a +ve charge +q. A positive charge +q is held at the point of support.
Then the time period of the bob is – [where, L = length of pendulum, geff = effective value of g]

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 19

Test: BITSAT Past Year Paper- 2009 - Question 20

Two tuning forks A and  B sounded together give 6 beats per second. With an air resonance tube closed at one end, the two forks give resonance when the two air columns are 24 cm and 25 cm respectively. Calculate the frequencies of forks.

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 20

Let the frequency of the first fork be f1 and that of second be f2.
We then have,

We also see that f1 > f2

Solving (i) and (ii), we get
f1 = 150 Hz   and f2 = 144 Hz

Test: BITSAT Past Year Paper- 2009 - Question 21

If an electron has an initial velocity in a direction different from that of an electric field, the path of the electron is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 21

The path is a parabola, because initial velocity can be resolved into two rectangular components, one along  and other ⊥ to  . The former decreases at a constant rate and latter is unaffected. The resultant path is therefore a parabola.

Test: BITSAT Past Year Paper- 2009 - Question 22

If on combining two charged bodies, the current does not flow then

Test: BITSAT Past Year Paper- 2009 - Question 23

Calculate the area of the plates of a one farad parallel plate capacitor if separation between plates is 1 mm and plates are in vacuum

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 23

For a parallel plate capacitor


This corresponds to area of square of side 10.6 km which shows that one farad is very large unit of capacitance.

Test: BITSAT Past Year Paper- 2009 - Question 24

The length of a potentiometer wire is ℓ. A cell of emf E is balanced at a length ℓ/3 from the positive end of the wire. If the length of the wire is increased by ℓ/2. At what distance will be the same cell give a balance point.

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 24

Potential gradient in the first case 

Potential gradient in second case

From equations (i) and (ii),

Test: BITSAT Past Year Paper- 2009 - Question 25

A conducting circular loop  of radius r carries a constant current i. It is placed in a uniform magnetic field such that is perpendicular to the plane of the loop. The magnetic force acting on the loop is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 25

The magnetic field is perpendicular to the plane of the paper. Let us consider two diametrically opposite elements. By Fleming's Left hand rule on element AB the direction of force  will be Leftwards and the magnitude will be
dF = Idl B sin 90° = IdlB

On element CD, the direction of force will be towards right on the plane of the papper and the magnitude will be dF = IdlB.

Test: BITSAT Past Year Paper- 2009 - Question 26

An ammeter reads upto 1 ampere. Its internal resistance is 0.81ohm. To increase the range to 10 A the value of the required shunt is

Test: BITSAT Past Year Paper- 2009 - Question 27

At the magnetic north pole of the earth, the value of horizontal component of earth’s magnetic field and angle of dip are, respectively

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 27

At the magnetic north pole, the magnetic needle will point vertically. There is no component of earth’s magnetic field in the horizontal direction and the angle of dip (the angle  that the resultant magnetic field at the place makes with the horizontal) is 90°.
H = 0,   δ = 90° (maximum)

Test: BITSAT Past Year Paper- 2009 - Question 28

Lenz’s law is a consequence of the law of conservation of

Test: BITSAT Past Year Paper- 2009 - Question 29

The instantaneous current from an a.c. source is I = 6 sin 314 t. What is the rms value of the current?

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 29

Test: BITSAT Past Year Paper- 2009 - Question 30

A coil has resistance 30 ohm and inductive reactance 20 ohm at 50 Hz frequency. If an ac source, of 200 volt, 100 Hz, is connected across the coil, the current in the coil will be

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 30

If ω = 50 × 2π then wL = 20Ω If ω' = 100 × 2π then ω'L = 40Ω Current flowing in the coil is

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