Physics - Online Test

Under Graduate Entrance Exams NEET Entrance Exam Physics TEST : 116

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

Q1.

A jet plane is travelling towards west at a speed of 1800 km/h. If the Earth’s magnetic field at the location has a magnitude of $5 \times 10 - 4 T$ and the dip angle is $3 0^{\circ}$ , voltage difference developed between the ends of the wing having a span of 25 m is approximately

A. 3.3 V

B. 3.7 V

C. 3.5 V

D. 3.1 V .

Answer : Option D
Explaination / Solution:

$e = V v l$

V = Vertical component of magnetic field

$e = ((B \sin ϕ) v l = 5 \times 10^{- 4} \times \frac{1}{2}) \times 500 \times 25 = 3.1 V$

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

Q2.

Suppose that the electric field part of an electromagnetic wave in vacuum is E = {(3.1 N/C) cos [(1.8 rad/m) y + (5.4

\times 1 0^{6}

rad/s)t]} i . Wavelength

λ

, frequency ν and the amplitude of the magnetic field part of the wave are

A. 5.5 m, 96 MHz, 100 nT

B. 4.0 m, 86 MHz, 250 nT

C. 3.5 m, 90 MHz, 200 nT

D. 3.5 m, 86 MHz, 100 nT

Answer : Option D
Explaination / Solution:
No Explaination.

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

Q3.

When an electron enters a magnetic field of 0.01 T with a speed of

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

it describes a circle of radius 6 mm there. Then specific charge of the electron is given by

A. 1.57

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

B. 1.67

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

C. 1.77

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

D. 1.87

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

Answer : Option B
Explaination / Solution:

evB=mv2rem=vrB=1076×10−3×0.01=1.67×1011CKg−1

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

Q4. Sky wave propagation is possible in the range from

A. none of these

B. few MHz up to 30 to 40 MHz

C. few Hz up to 30 to 40 kHz

D. few kHz up to 3.0 to 4.0 MHz

Answer : Option B
Explaination / Solution:
No Explaination.

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

Q5. A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?

A. 0.84 m

s^{- 2}

, 58.5

^{\circ}

with the direction of velocity

B. 0.82 m

s^{- 2}

, 59.5

^{\circ}

with the direction of velocity

C. 0.86 m

s^{- 2}

, 54.5

^{\circ}

with the direction of velocity

D. 0.85 m

s^{- 2}

, 56.5

^{\circ}

with the direction of velocity

Answer : Option C
Explaination / Solution:

Speed of the cyclist, v = 27 km/h = 7.5 m/s
Radius of the circular turn, r = 80 m
Centripetal acceleration is given as:
ac = $\frac{v^{2}}{r} = \frac{(7.5)^{2}}{80} =$ 0.7 ms-2

The situation is shown in the given figure:

Suppose the cyclist begins cycling from point P and moves toward point Q. At point Q, he applies the breaks and decelerates the speed of the bicycle by 0.5 m/s2.

This acceleration is along the tangent at Q and opposite to the direction of motion of the cyclist.
Since the angle between ac and aT is 900, the resultant acceleration a is given by:
a = $\sqrt{(a_{c})^{2} + (a_{T})^{2}} = \sqrt{(0.7)^{2} + (0.5)^{2}}$
= $\sqrt{0.74}$ = 0.86 ms-2

tan θ = $\frac{a_{c}}{a_{T}}$
where θ is the angle of the resultant with the direction of velocity.
tan θ = $\frac{0.7}{0.5}$ = 1.4
θ = tan-1(1.4) = 54.56° with the direction of velocity

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

Q6. Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is

6.5 \times 1 0^{- 7} C

The radii of A and B are negligible compared to the distance of separation.

A. 3.5×10−2N

B. 1.5×10−2N

C. 4.5×10−2N

2.5 \times 1 0^{- 2} N

Answer : Option B
Explaination / Solution:

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

Q7.

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

c m^{3}

what is the bulk modulus of the material?

A. 5.0

\times

1 0^{9}

B. 4.8

\times

1 0^{9}

C. 4.4

\times

1 0^{9}

D. 4.6

\times

1 0^{9}

Answer : Option B
Explaination / Solution:
No Explaination.

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

Q8.

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. How much work is done by the gas during the expansion?

A. 2.8

\times

1 0^{3}

B. 2.7

\times

1 0^{3}

C. 3.0

\times

1 0^{3}

D. 2.9

\times

1 0^{3}

Answer : Option B
Explaination / Solution:

In isothermal process

$W = n R T \log_{e} (\frac{V_{f}}{V_{i}})$

$W = 1 \times 8.31 \times 273 \times 2.303 \log_{10} \frac{10}{3} = 2.7 \times 10^{3} J$

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

Q9. A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

A. Lyman series: 143 nm and 122 nm; Balmer series: 646 nm.

B. Lyman series: 133 nm and 123 nm; Balmer series: 656 nm.

C. Lyman series: 103 nm and 122 nm; Balmer series: 656 nm.

D. Lyman series: 153 nm and 125 nm; Balmer series: 556 nm.

Answer : Option C
Explaination / Solution:
No Explaination.

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

Q10.

A man of mass 70 kg stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 5 m $s^{- 2}$ what would be the reading on the scale?

A. 150 kg

B. 130 kg

C. 120 kg

D. 105 kg

Answer : Option D
Explaination / Solution:

When the lift moves upward with acceleration = 5 ms-2 the net force acting upward

$\begin{aligned} R - m g = m a \\ R = m g + m a \\ R = m (g + a) \\ R = 70 (10 + 5) \\ R = 1050 N \end{aligned}$

R = 1050 N

(We experience weight due to reaction)

therefore Apparent weight = $= \frac{1050}{g} = \frac{1050}{10} = 105 k g$

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Under Graduate Entrance Exams NEET Entrance Exam Physics TEST : 116