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BITSAT Practice Test - 1 - JEE MCQ


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30 Questions MCQ Test BITSAT Mock Tests Series & Past Year Papers 2025 - BITSAT Practice Test - 1

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BITSAT Practice Test - 1 - Question 1

An electric motor operates at 20 r.p.s. What will be the approximate power delivered by the motor, if it supplies a torque of 75 N-m?

Detailed Solution for BITSAT Practice Test - 1 - Question 1

To find the power delivered by the motor, we can use the formula:

Power = Torque × Angular speed

First, we need to convert the rotational speed from r.p.s (revolutions per second) to radians per second. There are 2π radians in one revolution, so:

Angular speed = 20 r.p.s × 2π radians/revolution = 40π radians/second

Now, we can plug the values into the formula:

Power = 75 N-m × 40π radians/second = 3000π Watts

Approximating π as 3.14, we get:

Power ≈ 3000 × 3.14 Watts ≈ 9420 Watts

So, the approximate power delivered by the motor is 9420 Watts.

BITSAT Practice Test - 1 - Question 2

The main source of solar energy is

Detailed Solution for BITSAT Practice Test - 1 - Question 2

Main Source of Solar Energy: Fusion Reactions

Solar energy, which is harnessed from the sun, primarily comes from fusion reactions that occur in the sun's core. Fusion reactions involve the combining of two light atomic nuclei to form a heavier nucleus, releasing a large amount of energy in the process. This is the process that powers the sun and provides the Earth with the energy we receive in the form of sunlight.

Key Points:

- Fusion reactions in the sun's core involve the fusion of hydrogen atoms to form helium.
- These reactions release a tremendous amount of energy in the form of heat and light.
- The energy produced in the sun's core is radiated outwards in the form of sunlight.
- Solar panels on Earth capture this sunlight and convert it into usable electricity for various applications.

Overall, fusion reactions in the sun are the main source of solar energy that we harness and utilize for various purposes on Earth.

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BITSAT Practice Test - 1 - Question 3

The ionisation potential of H-atom is 13.6V. When it is excited from ground state by monochromatic radiations of 970.6 Å, the number of emission lines will be (according to Bohr's theory)

Detailed Solution for BITSAT Practice Test - 1 - Question 3

BITSAT Practice Test - 1 - Question 4

A piece of Aluminium and Germanium each are cooled from T1 k to T2 k. The resistance of

Detailed Solution for BITSAT Practice Test - 1 - Question 4

Explanation:

Thermal Coefficient of Resistance:
- The resistance of a material changes with temperature due to the temperature coefficient of resistance.
- For most metals like Aluminum, the resistance decreases as temperature increases.
- For semiconductors like Germanium, the resistance increases as temperature increases.

Cooling from T1 to T2:
- When both Aluminum and Germanium are cooled from T1 to T2, their resistances will change according to their respective temperature coefficients.
- Aluminum's resistance will decrease as it is a metal, and its resistance decreases with decreasing temperature.
- Germanium's resistance will increase as it is a semiconductor, and its resistance increases with decreasing temperature.

Conclusion:
- Therefore, the correct option is C: Aluminum decreases and that of Germanium increases.

BITSAT Practice Test - 1 - Question 5

Three resistors are connected to form the sides of a triangle ABC, the resistance of the sides AB, BC and CA are 40 ohms, 60 ohms and 100 ohms respectively. The effective resistance between the points A and B in ohms will be

Detailed Solution for BITSAT Practice Test - 1 - Question 5
Here resistor of side BC & AC are in series so
equ resis.= 100+60
= 160Ω
now this resis.is in parallel with resistance of AB
so1/rₑ= (1/40)+(1/160)
equ.resistance = 32
BITSAT Practice Test - 1 - Question 6

Which of the following statements is correct

Detailed Solution for BITSAT Practice Test - 1 - Question 6
i) Photocell depends upon the amount of light falls on it, not on the frequency.

ii) The photocurrent is proportional to the intensity of incident light, not on the applied voltage.

iii) Photoelectric current increases with an increase in intensity.

iv) Stopping potential increases with increase in Kinetic energy which depends upon the frequency, not on the intensity.

So, the correct answer is (c).
BITSAT Practice Test - 1 - Question 7

Threshold wavelength for photoelectric emission from a metal surface is 5200 Å. Photoelectrons will be emitted when this surface is illuminated with monochromatic radiation from

Detailed Solution for BITSAT Practice Test - 1 - Question 7
Ans : (B)
For photoelectric emission to take place the wavelength of incident
light must be less than
threshold value which is 5200 A∘
wavelength of IR light = 7800 A∘
wavelength of UV light = 4000 A∘
BITSAT Practice Test - 1 - Question 8

The number of turns of primary and secondary coils of a transformer are 5 and 10 respectively and mutual inductance of the transformer is 25 H. Now, number of turns in primary and secondary are made 10 and 5 respectuvely. Mutual inductance of transformer will be

Detailed Solution for BITSAT Practice Test - 1 - Question 8
BITSAT Practice Test - 1 - Question 9

The capacitor of a parallel plate capacitor with no dielectric substance and a separation of 0.4 cm is 2 μF. The separation is reduced to half and it is filled with a substance of dielectric value 2.8. The new capacity of the capacitor is

Detailed Solution for BITSAT Practice Test - 1 - Question 9

We know that capacitance, c = Kε0A / d
Thus we get when K = 1 and d = 0.4 we get c = 2F
Thus let the value of c when K = 2.8 and d = 0.2 be z
Thus we get z/2 = (1 / 0.4) / (2.8 / 0.2) 
Thus we get z = 2 x 5.6 F
= 11.2 F

BITSAT Practice Test - 1 - Question 10

Two concentric uniformly charged spheres of radius 10cm and 20cm are arranged as shown in the figure. Potential difference between the spheres is:

Detailed Solution for BITSAT Practice Test - 1 - Question 10

BITSAT Practice Test - 1 - Question 11

A block of mass 10 kg is placed on an inclined plane. When the angle of inclination is 30 o, the block just begins to side down the plane. The force of static friction is

Detailed Solution for BITSAT Practice Test - 1 - Question 11


Force of static friction f = μR
or f = mg sin θ, on point of sliding down of body.
∴ f = 10 x g x sin 30o
= 5(g) N = 5 kg wt.

BITSAT Practice Test - 1 - Question 12

The angular velocity of rotation of star at which the matter start to escape from its equator will be

Detailed Solution for BITSAT Practice Test - 1 - Question 12

BITSAT Practice Test - 1 - Question 13

A perfect black body emits radiation at temperature 300 K. If it radiates 16 times this amount, then its temperature will be

Detailed Solution for BITSAT Practice Test - 1 - Question 13

BITSAT Practice Test - 1 - Question 14

One litre of Helium gas at a pressure 76 cm of Hg and temperture 27o is heated till its pressure and volume are doubled. The final temperature attained by the gas is

Detailed Solution for BITSAT Practice Test - 1 - Question 14
P1V1/T1=4P1V1/T2
T1=300K
THEN,
1/300=4/T2
T2=1200K
= 927degreeC
BITSAT Practice Test - 1 - Question 15

If current in a moving coil galvanometer is increased by 1%, then percentage increase in its deflection will be

Detailed Solution for BITSAT Practice Test - 1 - Question 15

Explanation:

- When the current in a moving coil galvanometer is increased by 1%, the magnetic field strength created by the current also increases by 1%.
- The deflection of a moving coil galvanometer is directly proportional to the magnetic field strength.
- Therefore, an increase of 1% in current will result in a 1% increase in the deflection of the galvanometer.

Therefore, the percentage increase in deflection will be 1%. Hence, option A is correct.

BITSAT Practice Test - 1 - Question 16

A bar magnet is held perpendicular to a uniform magnetic field. If the couple acting on the magnet is to be halved by rotating it, then the angle by which it is to be rotated is

Detailed Solution for BITSAT Practice Test - 1 - Question 16

BITSAT Practice Test - 1 - Question 17

What should be the minimum average velocity of water in a small tube of radius 5 mm, so that the flow is just turbulent? (Given: Viscosity of water=0.01 poise and density of water = 1000 kg-m⁻3)

Detailed Solution for BITSAT Practice Test - 1 - Question 17

BITSAT Practice Test - 1 - Question 18

The neck and bottom of a bottle are 3 cm and 15 cm in radius respectively. If the cork is pressed with a force 12 N in the neck of the bottle, then force exerted on the bottom of the bottle is

Detailed Solution for BITSAT Practice Test - 1 - Question 18

99996237221-0-0

BITSAT Practice Test - 1 - Question 19

Two similar wires under the same load yield elongations of 0.1 mm and 0.05 mm respectively. If the area of cross-section of the first wire is 4 mm2, then the area of cross-section of the second wire is

Detailed Solution for BITSAT Practice Test - 1 - Question 19

BITSAT Practice Test - 1 - Question 20

A stone is released from the top of a tower. If its velocity at half of the height is 10 m−s−1, then height of the tower is (Take g = 10 m−s−2)

Detailed Solution for BITSAT Practice Test - 1 - Question 20
V² = 2as, = (10)² = 2*10*h/2 , ,,,,, and solve it, then, h = 10 m. Ans.
BITSAT Practice Test - 1 - Question 21

For a particle in a uniformly accelerated circular motion

Detailed Solution for BITSAT Practice Test - 1 - Question 21



 

Explanation:

 


  • For a particle in a uniformly accelerated circular motion, the velocity is transverse to the radius of the circle. This means that the velocity is perpendicular to the radius of the circle at any point in time.

  • The acceleration of the particle in this motion has both radial and transverse components. The radial component of acceleration is directed towards the center of the circle, while the transverse component is perpendicular to the velocity and directed tangentially to the circle.

  • Therefore, in a uniformly accelerated circular motion, the velocity is transverse and the acceleration has both radial and transverse components.


  •  
BITSAT Practice Test - 1 - Question 22

The resultant of two forces acting at an angle of 150o is 10 N and is perpendicular to one of the forces. The other force is

Detailed Solution for BITSAT Practice Test - 1 - Question 22

BITSAT Practice Test - 1 - Question 23

A body weighs 8 gm when placed in one pan and 18 gm when placed on the other pan of a false balance. If the beam is horizontal when both the pans are empty, the true weight of the body is

Detailed Solution for BITSAT Practice Test - 1 - Question 23

If the beam is horizontal, the apparent weight 

BITSAT Practice Test - 1 - Question 24

The length of a telescope is 36 cm. The focal lengths of its lenses can be

Detailed Solution for BITSAT Practice Test - 1 - Question 24
BITSAT Practice Test - 1 - Question 25

A mass M is suspended from the spring of negligible mass. The spring is pulled a little and then released, so that the mass executes simple harmonic motion of time period T. If the mass is increased by m, the time period becomes 5T/3. The ratio m/M is

Detailed Solution for BITSAT Practice Test - 1 - Question 25
BITSAT Practice Test - 1 - Question 26

A body esecutes S.H.M. with amplitude A₀. At what displacement from the mean position, potential energy is one-fourth of its total energy?

Detailed Solution for BITSAT Practice Test - 1 - Question 26

Solution:

- Given that potential energy is one-fourth of total energy when displacement from the mean position is x.
- Total energy in S.H.M is the sum of kinetic energy and potential energy.
- At any point in S.H.M, total energy E = K.E + P.E = 1/2 kx² + 1/2 kA₀²
- Given that at the point where potential energy is one-fourth of total energy, P.E = 1/4 E
- Therefore, 1/2 kx² = 1/4 (1/2 kx² + 1/2 kA₀²)
- Simplifying the equation, we get 2x² = x² + A₀²
- Therefore, x² = A₀²
- Taking square root on both sides, x = A₀
- Hence, the displacement from the mean position where potential energy is one-fourth of its total energy is A₀.

Therefore, the correct answer is B: A₀/2.

BITSAT Practice Test - 1 - Question 27

A compound microscope has two lenses. The magnifying power of one is 5 and the combined magnifying power is 100. The magnifying power of the other lens is

Detailed Solution for BITSAT Practice Test - 1 - Question 27

BITSAT Practice Test - 1 - Question 28

A cylindrical solid of mass M has radius R and length L. Its moment of inertia about a generator is

Detailed Solution for BITSAT Practice Test - 1 - Question 28

Correct Answer : d

Explanation :  Generator of axis of a cylinder is a line lying on its surface and parallel to axis of cylinder

By parallel axis theorem

I = (MR2 ) /2 + MR2

= 3/2MR2

BITSAT Practice Test - 1 - Question 29

Which of the following statements are correct when a gymnast sitting on a rotating stool with his arms outstretched, suddenly lowers his hands?

1. The angular velocity decreases.
2. The angular velocity increases.
3. The moment of inertia decreases.
4. The moment of inertia increases.
Select the answers using the codes given below.

 

Detailed Solution for BITSAT Practice Test - 1 - Question 29

Explanation:


 


  • When the gymnast lowers his hands while sitting on a rotating stool, the moment of inertia decreases. This is because the distribution of mass is closer to the axis of rotation, leading to a decrease in moment of inertia.

  • As the moment of inertia decreases, according to the conservation of angular momentum, the angular velocity of the gymnast increases. This is because angular momentum is conserved and is the product of moment of inertia and angular velocity.


  •  



Therefore, the correct statements are:
2. The angular velocity increases.
3. The moment of inertia decreases.

BITSAT Practice Test - 1 - Question 30

Two identical particles move towards eachother with velocity 2ν and ν respectively. The velocity of centre of mass is

Detailed Solution for BITSAT Practice Test - 1 - Question 30
M(2v)-m(v)/m+m = v/2
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