Physics | NEET Entrance Exam | Under Graduate Entrance Exams Online Objective Test

Under Graduate Entrance Exams NEET Entrance Exam # Physics Dual Nature of Radiation and Matter Prepare / Learn

Q1. If an electron accelerated through a potential difference of 500 volt attains a speed of 1.33

\times 1 0^{7} m s^{- 1}

then specific charge of the electron should be

A. 1.86

\times 1 0^{11} C k g^{- 1}

B. 1.76

\times 1 0^{11} C k g^{- 1}

C. 1.66

\times 1 0^{11} C k g^{- 1}

D. 1.96

\times 1 0^{11} C k g^{- 1}

Answer : Option B
Explaination / Solution:

eV0=12mv2em=v22V0=(1.33×107)22×500=1.76×1011CKg−1

Under Graduate Entrance Exams NEET Entrance Exam # Physics Communication Systems Prepare / Learn

Q2. To increase the range of line of sight communication

A. decrease the height of transmitting tower only

B. increase the height of both transmitting and receiving towers

C. decrease the height of both transmitting and receiving towers

D. decrease the height of receiving tower only

Answer : Option B
Explaination / Solution:
No Explaination.

Under Graduate Entrance Exams NEET Entrance Exam # Physics Motion in A Plane Prepare / Learn

Q3. A cricketer hits a cricket ball from the ground so that it goes directly upwards. If the ball takes, 10 s to return to the ground, determine its maximum height.

A. 132.5 m

B. 112.5 m

C. 152.5 m

D. 122.5 m

Answer : Option D
Explaination / Solution:

We know that at the maximum height, the velocity of the ball is 0 m/s.

We also know that the ball takes the same time to reach its maximum height as it takes to travel from its maximum height to the initial position, due to time symmetry. The time taken is half the total time.

Therefore, we have the following information for the second (downward) part of the motion of the ball:

t = 5 second; half the total time

$v_{t o p} = 0 m / s$

g = 9.8 m/s2 downward

s = ?

As we know

s = ut+ $\frac{1}{2} g t^{2}$

Put all the given value,

=> s = $0 \times 5 + \frac{1}{2} \times 9.8 \times 25$

=> s = 122.5 m

Under Graduate Entrance Exams NEET Entrance Exam # Physics Electric Charges and Fields Prepare / Learn

Q4. Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is

8 .0 \times 1 0^{3} N m^{2} / C

. (a) What is the net charge inside the box? (b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box?

A. 0.05

μ C

, No

B. 0.04

μ C

, Yes

C. 0.07

μ C

, No

D. 0.06

μ C

, Yes

Answer : Option C
Explaination / Solution:

(a) Net outward flux through the surface of the box, ϕ = 8.0 ×

10^{3}

m^{2}

/C For a body containing net charge q, flux is given by the relation,

\in_{0}

= Permittivity of free space = 8.854 ×

10^{- 12}

N^{−1}

C^{2}

m^{- 2}

We have

ϕ = \frac{q}{\in_{0}} s o

q =

\in_{0}

ϕ = 8.854 ×

10^{- 12}

× 8.0 ×

10^{3}

= 7.08 ×

10^{- 8}

= 0.07 μC Therefore, the net charge inside the box is 0.07 μC. (b) No Net flux piercing out through a body depends on the net charge contained in the body. If net flux is zero, then it can be inferred that net charge inside the body is zero. The body may have equal amount of positive and negative charges.

Under Graduate Entrance Exams NEET Entrance Exam # Physics Mechanical Properties of Solids Prepare / Learn

Q5. A copper cube measures 6.00 cm on each side. The bottom face is held in place by very strong glue to a flat horizontal surface, while a horizontal force F is applied to the upper face parallel to one of the edges. How large must F be to cause the cube to deform by 0.250 mm? (shear modulus of copper = 4.4

\times

1 0^{10}

Pa)

A. 6.6

\times

1 0^{5}

B. 6.5

\times

1 0^{5}

C. 6.4

\times

1 0^{5}

D. 6.3

\times

1 0^{5}

Answer : Option A
Explaination / Solution:
No Explaination.

Under Graduate Entrance Exams NEET Entrance Exam # Physics Thermodynamics Prepare / Learn

Q6.

A 1.0-kg bar of copper is heated at atmospheric pressure. If its temperature increases from 20 $^{\circ}$ C to 50 $^{\circ}$ C, what is the work done by the copper on the surrounding atmosphere? Given that for Cu $3 α$ = 5.1 x 10-5 / C, $ρ_{C u} = 8.92 \times 10^{3} K g / m^{3}$

ρ_{C u} = 8.92 \times 10^{3} K g / m^{3}

A. 1.6

\times

1 0^{- 2}

B. 1.9

\times

1 0^{- 2}

C. 1.8

\times

1 0^{- 2}

D. 1.7

\times

1 0^{- 2}

Answer : Option D
Explaination / Solution:

ΔV=3αViΔT=5.1×10−5×30×Vi=1.5×10−3ViΔV=1.5×10−3mρCu=1.5×10−318.92×103=1.7×10−7W=PΔV=1.01×105×1.7×10−7=1.7×10−2J

Under Graduate Entrance Exams NEET Entrance Exam # Physics Laws of Motion Prepare / Learn

Q7.

A man of mass 70 kg stands on a weighing scale in a lift which is moving upwards with a uniform speed of 10 m $s^{- 1}$ . What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

A. 105 kg

B. 0 kg

C. 35 kg

D. 70 kg

Answer : Option B
Explaination / Solution:

When the lift falls freely under gravity, a=g

Therefore Apparent weight,

$R = m (g - a) = m (g - g) = 0$

This is the condition of weightlessness.

Under Graduate Entrance Exams NEET Entrance Exam # Physics Current Electricity Prepare / Learn

Q8. In the measurement of resistance by a metre bridge, the current is necessarily reversed through the bridge wire to eliminate

A. Index errors

B. Random errors

C. Error due to thermoelectric effects

D. End errors

Answer : Option D
Explaination / Solution:

The connecting metal strips include a small resistance in the circuit called the end resistance. On reversing the current, the end resistance tends to get cancelled out.

Under Graduate Entrance Exams NEET Entrance Exam # Physics Work Energy and Power Prepare / Learn

Q9. A block having a mass of 0.80 kg is given an initial velocity 1.2 m/s to the right and collides with a spring of negligible mass and force constant k =50 N/m. Assuming the surface to be frictionless, calculate the maximum compression of the spring after the collision.

A. 0.25 m

B. 0.10 m

C. 0.15m

D. 0.20 m

Answer : Option C
Explaination / Solution:

for maximum compression

$\begin{aligned} \frac{1}{2} k x^{2} = \frac{1}{2} m v^{2} \\ x = \sqrt{\frac{m v^{2}}{k}} = \sqrt{\frac{0.8 \times 1.2 \times 1.2}{50}} = 0.15 m \end{aligned}$

the maximum compression of the spring after the collision = 0.15m

Under Graduate Entrance Exams NEET Entrance Exam # Physics Atoms Prepare / Learn

Q10. Average angle of deflection of α -particles by a thin gold foil predicted by Thomson’s model is

A. incomparable to Rutherford’s model

B. about the more than predicted by Rutherford’s model

C. about the same as predicted by Rutherford’s model

D. about the less than predicted by Rutherford’s model

Answer : Option C
Explaination / Solution:
No Explaination.

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