NEET PHYSICS

DAILY PHYSICS TEST QUESTIONS

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1 Alternating Currents

A transformer having efficiency of 90% is working on 200V and 3kW power supply. If the current in the secondary coil is 6A, the voltage across the secondary coil and current in the primary coil respectively are

1) 300V,15A
2) 450V,15A
3) 450V,13.5A
4) 600V,15A

Option 1

Option 2 Solution:

Solution

We know efficiency $$\displaystyle\eta=\frac{{{O}{u}{t}{p}{u}{t}}}{{{I}{n}{p}{u}{t}}}=\frac{{{V}_{{{s}}}{I}_{{{s}}}}}{{{V}_{{{p}}}{I}_{{{p}}}}}$$

$$\displaystyle\Rightarrow\eta=\frac{{{V}_{{{s}}}{I}_{{{s}}}}}{{{V}_{{{p}}}{I}_{{{p}}}}}\Rightarrow{0.9}=\frac{{{V}_{{{s}}}{\left({6}\right)}}}{{{3}\times{10}^{{{3}}}}}$$

⇒Vs=450V
As VpIp=3000

$$\displaystyle{I}_{{{p}}}=\frac{3000}{{{V}_{{{p}}}}}=\frac{3000}{{200}}{A}={15}{A}$$

Option 3

Option 4

2 Magnetic Effects of Current and Magnetism

A steady current flows in a long wire. It is bent into a circular loop of one turn and the magnetic field at the centre of the coil is BB. If the same wire is bent into a circular loop of nn turns, the magnetic field at the centre of the coil is:

1) $$\dfrac{B}{n}$$
2) $$nB$$^?
3) $$nB^2$$
4) $$n^2B$$

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

Let the length of the wire be $$L$$ and radius of the coil be $$r$$

$$\therefore$$ $$L = 2\pi r$$

Magnetic field at the centre of coil $$B = \dfrac{\mu_o I}{2r}$$ ........(1)

Now the same wire is bent into circular loops of n turns.

$$\therefore$$ $$L =n 2\pi r' $$

$$2\pi r = 2n\pi r'$$ $$\implies r' = \dfrac{r}{n}$$ ........(2)

Magnetic field at the centre of circular loops of n turns $$B' = \dfrac{\mu_o I}{4\pi r'} \times 2\pi n = \dfrac{\mu_o I}{2r'}n$$

Using equation (2), we get $$B' = \dfrac{\mu_o I}{2 \frac{r}{n}}n = \dfrac{\mu_o I}{2 r} n^2$$

$$\implies$$ $$B' = n^2 B$$

3 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

4 Thermodynamics

A spherical body of emissivity e = 0.6, placed inside a perfectly black body is maintained at

temperature T then the energy radiated per second will be

(1)E = 0.4 eAT⁴ (2) E = 0.8 eAT⁴

(3) E = 0.6 eAT⁴ (4) E = 1.0eAT⁴

Option 1

Option 2

Option 3 Solution:

Option 4

5 Atoms and Nuclei

An electron jumps from 6th shell in an hydrogen atom, the number of Lyman emission lines will be:

1) 5
2) 6
3) 10
4) 12

Option 1 Solution:

Solution

From level 6 to level 1 i,e,

6→1

and 5→1

4→1

3→1

2→1

So total 5 lines in Lyman series.

Option 2

Option 3

Option 4

6 Electrostatics

A sphere carrying a charge of 0.01 C is kept at rest without falling down, touching a wall by applying an electric field 100 N/C. If the coefficient of friction between the sphere and the wall is 0.2, the weight of the sphere is:

1) 2 kg
2) 2 N
3) 20 N
4) 0.2 N

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

As the sphere is at rest N=Eq and f =mg
N = 100×0⋅01
N = 1Newton
we know that f=μN (maximum static function is μN)
f = 0⋅2×1
f=0⋅2
Condition of equlibrium:

f = Mg

⇒0⋅2=mg
∴ weight =0⋅2N
∴ Maximum weight which can be stopped is 0⋅2N
∴ weight must be ≤0⋅2N
From the given options , only( 4) is ≤0.2

X-rays are :

(1) sound waves (2) em waves

(3) radio waves (4) all above

Option 1

Option 2 Solution:

Option 3

Option 4

8 Alternating Currents

An ideal inductor is in turn put across 220 V. 50 Hz and 220 V, 100 Hz supplies. The current flowing through it in the two cases will be then

1) equal
2) different
3) zero
4) infinite

Option 1

Option 2 Solution:

Solution

The current in the inductor coil is given by $$I \, = \, \dfrac{V}{X_L} \, = \, \dfrac{V}{2\pi \upsilon L}$$
Since frequency $$\upsilon$$ in the two cases is different, hence the current in two cases will be different.

Option 3

Option 4

9 Electromagnetic Induction

A source of 220V is applied in an A.C. circuit. The value of resistance is 220Ω. Frequency & inductance are 50 Hz and 0.7 H, then wattless current is

1) 0.5amp
2) 0.7amp
3) 1.0amp
4) none

Option 1

Option 2 Solution:

Solution

$$ \Rightarrow R = 2\Omega $$

$$ \Rightarrow {V_{rms}} = 220V$$

$$ \Rightarrow {V_ \circ } = 220\sqrt 2 $$

$$\Rightarrow {X_L} = WL = 100\pi \times 0.7$$

$$ \Rightarrow 100\dfrac{{22}}{7} \times \dfrac{7}{{10}} = 220\Omega $$

$$\therefore z = 220\sqrt 3 $$

also $$\tan \phi = \dfrac{{{V_L}}}{R} = 1$$

$$\phi = {45^ \circ }$$

$$\therefore$$ wattless current$$ = {i_ \circ }\sin \phi $$

$$ = \dfrac{{220\sqrt 3 }}{{220\sqrt 2 }}\sin {45^ \circ }$$

$$ = \dfrac{1}{{\sqrt 2 }} = 0.7A$$

so we get $$0.7A$$

Option 3

Option 4

10 Magnetic Effects of Current and Magnetism

At the mid-point along the length of a long solenoid, the magnetic field is equal to x, If the length of the solenoid is doubled and the current is reduced to half, the field at the new mid-point will be nearest to:

(1) x/2
(2) x/4
(3) x
(4) 2x

Option 1

Option 2 Solution:

Solution

x/4

Option 3

Option 4

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