NEET PHYSICS

DAILY PHYSICS TEST QUESTIONS

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1 Physical world and measurement

The pitch of a screw gauge is 1mm and there are 100 divisions on circular scale. When faces A and B are just touching each without putting anything between the studs 32nd division on the circular scale coincides with reference line. When a glass plate is placed between the studs, the linear scale reads 4 divisions and the circular scale reads 16 divisions. Find the thickness of the glass plate. Zero of linear scale is not hidden from circular scale when A and B touches each other.

1) 3.841mm
2) 3.84mm
3) 3.85mm
4) 3.846mm

Option 1

Option 2 Solution:

As zero is not hidden from cricular scale when A and B touches each other.

Hence, the screw gauge has positive error.
e=+n(LC)=32×0.01=0.32 mm
Linear scale reading =4×(1mm)=4mm
Circular scale reading =16×(0.01mm)=0.16mm
∴ Measured reading-e =(4+0.16)mm=4.16 mm
∴ Absolute reading = Measured reading-e
=(4.16−0.32)mm=3.84mm
Therefore, thickness of the glass plate is 3.84mm.

Option 3

Option 4

In a triode amplifier phase between input and output voltages is :

(1) zero (2) p

(3) 2 p (4) p/2

Option 1

Option 2 Solution:

Option 3

Option 4

3 Motion of System of Particles and Rigid Body

A mass is whirled in a circular path with a constant angular velocity and its angular momentum is L. If the length of string is now halved keeping the angular velocity the same, the new angular momentum is:

1) L /4
2) L /2
3) L
4) 2L

Option 1 Solution:

Solution

Angular momentum, $$L= mvr$$

Also $$v=r\omega \implies L= mr^2\omega$$

Now, $$r_2= \dfrac{r}{2}$$ (putting it in the earliar equation to get new momentum)

Thus $$ L_2= m\dfrac{r^2}{4}\omega$$

$$\implies L_2= \dfrac{L}{4}$$

Option 2

Option 3

Option 4

4 Electromagnetic Induction

A coil has an area of 0.05m² and it has 800 turns. It is placed perpendicular in a magnetic field of strength 4×10⁻⁵Wb/m², it is rotated through 90^o in 0.1sec. The average emf induced in the coil is

1) 0.056V
2) 0.046V
3) 0.026V
4) 0.016V

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

$$\phi_1 = NBA$$

$$\phi_2 = O$$

$$E_avg = \dfrac{NBA}{Ot}$$

$$= \dfrac{800 \times 4 \times 10^{-5} \times 0.05}{0.1}$$

= 0.016V

5 Gravitation

If the mass of the earth increases by 80%80% and radius of the earth increases by 40%40% then the percentage charge in acceleration due to gravity on the surface of radius of earth is (where $$g_{s}=\dfrac{GM} {R^{2}} ,M=$$ mass of earth and $$R=$$ radius of earth})

1) zero
2) +8.16%
3) - 8.16%
4) 160%

Option 1

Option 2

Option 3 Solution:

Solution

As per question final radius = 1.4 times initial radius

And final mass =1.8 times initial mass

Using proportionality we can write final g$$_s$$= $$\dfrac{1.8} {1.96} $$ times initial g$$_s$$

Thus acceleration due to gravity decrease by approximately 8.16%

Option 4

6 Magnetic Effects of Current and Magnetism

Two infinitely long straight conductors which are held parallel to each other at a distance 0.4m carry currents 20A and 10A in the same direction. The magnetic induction at a point midway between them is

1) 10⁻⁵T
2) 3×10⁻⁵T
3) 2×10⁻⁵T
4) 10⁻⁶T

Option 1 Solution:

Solution

$$\left | B \right |=(\dfrac{\mu _{0}i_{1}}{2\pi r_{1}})-(\dfrac{\mu _{0}i_{2}}{2\pi r_{2}})$$
=$$(\dfrac{2\times 10^{-7}\times 20}{0.2})-(\dfrac{2\times 10^{-7}\times 10}{0.2})$$
=$$50\times 2\times 10^{-7}T$$=$$10^{-5}T$$

Option 2

Option 3

Option 4

The temperature of 100 g of water is to be raised from 24� C to 90� C by adding steam to it. Calculate the mass of the steam required for this purpose (latent heat of steam = 540 k Cal/kg)
1.10 g
2. 11 g
3. 12 g
4. 15 g

Option 1

Option 2

Option 3 Solution:

Option 4

8 Wave Optics

Which of the following is correct ?

(1) Light and solid are both transverse waves.
(2) Coherent source of light passing through two slits causes interference.
(3) Interference of light is evidence for transverse character of light.
(4) Rainbow appears because of scattering of light.

Option 1

Option 2 Solution:

Option 3

Option 4

9 Dual Nature of Matter and Radiation

The Kα X-ray emission line of tungsten occurs at λ = 0.021nm. The energy difference between K and L levels in this atom is about

1) 0.51MeV
2) 1.2MeV
3) 59keV
4) 13.6eV

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

Line $$\displaystyle { K }_{ \alpha }$$ corresponds to transfer of electron from L-shell to K-shell.
Here, $$\displaystyle \lambda =0.021nm = 2.1\times 10^{-11}\ m$$

We know that $$\lambda = \dfrac{hc}{E (in \ Joule)}$$
$$\displaystyle \lambda=\frac { hc }{ E (in \ Joule)} =\frac { hc }{ e\times E\left( in \ eV \right) } $$
$$\displaystyle \Rightarrow \quad E\left(in \ eV \right) =\frac { hc }{ e\lambda } =\frac { 6.63\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 1.6\times { 10 }^{ -19 }\times 2.1\times { 10 }^{ -11 } } eV$$
$$\displaystyle E(in \ eV)=5.9\times { 10 }^{ 4 }eV=59keV$$

10 Alternating Currents

A 100 volt A.C. source of frequency 500 hertz is connected to a L-C-R circuit with L == 8.1 millihenry, C == 12.5 microfarad and R == 10 ohm, all connected in series. The potential difference across the resistance will be:

1) 10V
2) 100V
3) 50V
4) 500V

Option 1

Option 2 Solution:

Solution

Impedance of series RLC circuit is given by:

$$Z = \sqrt {R^2 + \left( 2 \pi f L - \dfrac{1}{2 \pi f C} \right)^2}$$

$$Z = \sqrt {10^2 + \left(2\pi \times 500 \times 8.1 \times10^{-3} - \dfrac{1}{2 \pi \times 500 \times 12.5 \times 10^{-6} } \right)^2}$$

$$Z \approx 10\ \Omega$$

Current flowing through the circuit is:

$$I = \dfrac{V}{Z} = 10A$$

Potential difference across the resistor, by ohm's Law is:

$$V_R = IR = 10 \times 10 = 100\ V$$

Option 3

Option 4

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