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

Under Graduate Entrance Exams NEET Entrance Exam # Physics Electromagnetic Waves Prepare / Learn

Q1. About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation at a distance of 1m from the bulb and at a distance of 10 m are

A. 0.44

W / m^{2}

, 0.004

W / m^{2}

B. 0.55

W / m^{2}

, 0.004

W / m^{2}

C. 0.5

W / m^{2}

, 0.004

W / m^{2}

D. 0.45

W / m^{2}

, 0.004

W / m^{2}

Answer : Option A
Explaination / Solution:
No Explaination.

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

Q2.

1.If an electron moving with a speed of 2.5

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

is deflected by an electric field of 1.6 k V

m^{- 1}

perpendicular to its circular path, then e/m for the electron will be (given radius of circlar path = 2.3 m)

A. 1.85

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

B. 1.9

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

C. 1.7

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

D. 1.8

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

Answer : Option C
Explaination / Solution:

Electric field provide required centripetal force for circular motion

$\begin{aligned} e E = \frac{m v^{2}}{r} \\ \frac{e}{m} = \frac{v^{2}}{r E} = \frac{{(2.5 \times 10^{7})}^{2}}{2.3 \times 1.6 \times 10^{3}} = 1.7 \times 10^{11} C K g^{- 1} \end{aligned}$

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

Q3. Space wave propagation can be

A. Line of sight communication or satellite communication

B. none of these

C. can only be transmitted from a space station

D. can only be transmitted from a polar satellite

Answer : Option A
Explaination / Solution:
No Explaination.

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

Q4.

A ball is thrown upwards with an initial velocity of 10 m $s^{- 1}$ . Determine the maximum height reached above the thrower’s hand. Determine the time it takes the ball to reach its maximum height.

A. 5.43 m, 0.92 s

B. 5.23 m, 1.12 s

C. 5.10 m, 1.02 s

D. 5.25 m, 0.42 s

Answer : Option C
Explaination / Solution:

Initial velocity u = 10m/s

As at the maximum height ball ll stop so final velocity v = 0 m/s

Only acceleration working on it is acceleration due to gravity g = -9.8m/s2

Let height = h

So we know

$v^{2} - u^{2} = 2 a s$

=> $(0)^{2} - (10)^{2} = 2 \times (- 9.8) h$

=> $h = \frac{- 100}{- 19.6}$ 5.10 m

Also let time taken to reach maximum height = t

Then

We know

v = u +at

=> 0 = 10 +(-9.8)t

=> t = $\frac{- 10}{- 9.8} = 1.02 s$

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

Q5. Consider a uniform electric field

E = 3 \times 1 0^{3} N / C

(a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a

6 0^{\circ}

angle with the x-axis?

A. 40Nm2/C,25Nm2/C

B. 20Nm2/C,15Nm2/C

C. 40Nm2/C,15Nm2/C

D. 30Nm2/C,15Nm2/C

Answer : Option D
Explaination / Solution:

(a) Electric field intensity, = 3 ×

10^{3}

î N/C Magnitude of electric field intensity, =

3 \times 10^{3}

N/C Side of the square, s = 10 cm = 0.1 m Area of the square, A =

s^{2}

= 0.01

m^{2}

The plane of the square is parallel to the y-z plane. Hence, angle between the unit vector normal to the plane and electric field,

θ

= 0° Flux (Φ) through the plane is given by the relation, ϕ =

\vec{E} . \vec{A} = E A C o s θ

=3 ×

10^{3}

× 0.01 × cos0° = 30 N

m^{2} / C

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

Q6. A specimen of oil having an initial volume of 600

c m^{3}

is subjected to a pressure increase of 3.6

\times

1 0^{6}

Pa and the volume is found to decrease by 0.45 cm3 what is the compressibility of the material?

A. 2.3

\times

1 0^{- 10}

^{- 1}

B. 2.1

\times

1 0^{- 10}

^{- 1}

C. 1.7

\times

1 0^{- 10}

^{- 1}

D. 1.9

\times

1 0^{- 10}

^{- 1}

Answer : Option B
Explaination / Solution:
No Explaination.

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

Q7. A 1.0-mol sample of an ideal gas is kept at 0.0

^{\circ}

C during an expansion from 3.0 L to 10.0 L. If the gas is returned to the original volume by means of an isobaric process, how much work is done by the gas?

A. −1.6×103J

B. -1.3

\times

1 0^{3}

C. -1.5

\times

1 0^{3}

D. -1.4

\times

1 0^{3}

Answer : Option A
Explaination / Solution:

$\begin{aligned} W = n R (T_{f} - T_{i}) \\ V \propto T \\ \frac{10}{273} = \frac{3}{T_{f}} \\ T_{f} = 81.9 K \\ T_{i} = 273 K \end{aligned}$

$W = 1 \times 8.31 \times (81.9 - 273) = - 1.6 \times 10^{3} J$

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

Q8. A man of mass 70 kg stands on a weighing scale in a lift which is freely falling under gravity. What would be the reading on the scale?

A. 95 kg Hz

B. 125 kg

C. 115 kg

D. 0 kg

Answer : Option D
Explaination / Solution:

When the lift fall freely under gravity, a = g

Therefore Apparent weight, (We experience weight due to reaction)

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

This is the condition for weightlessness.

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

Q9.

The Sun converts an enormous amount of matter to energy. Each second, 4.19 $\times$ $1 0^{9}$ kg—approximately the capacity of 400 average-sized cargo ships—is changed to energy. What is the power output of the Sun?

\times

2.62

\times

1 0^{26}

B. 1.57

\times

1 0^{26}

0.72

\times

1 0^{26}

D. 3.77×1026W

Answer : Option D
Explaination / Solution:

Energy liberated per second

$E = m c^{2} = 4.19 \times 10^{9} \times 3 \times 10^{8} \times 3 \times 10^{8} = 37.71 \times 10^{25} J$

power output of sun is equal to energy output per second

$P = \frac{W}{t} = \frac{37.71 \times 10^{25}}{1} = 3.77 \times 10^{26} W$

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

Q10. n accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5

\times 1 0^{11}

m with orbital speed 3

\times 1 0^{4}

m/s. (Mass of earth= 6.0

\times 1 0^{24}

kg.)

A. 2.7

\times 1 0^{74}

B. 2.9

\times 1 0^{74}

C. 2.6

\times 1 0^{74}

D. 2.8

\times 1 0^{74}

Answer : Option C
Explaination / Solution:
No Explaination.

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