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

A shot leaves a gun at the rate of 200 m/s. The greatest distance to which it could be projected is:
1. 200 m
2. 1000 m
3. 2000 m
4. 4000 m

Option 1

Option 2

Option 3 Solution:

Option 4

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 Capacitance

Two condensers of 1 mF are connected in series with a battery of 6 volt, the total charge on condensers will be :

(1) 2 mC
(2) 2.5 mC
(3) 9 mC
(4) 4 mC

Option 1 Solution:

Option 2

Option 3

Option 4

5 Dual Nature of Matter and Radiation

Assuming an average energy of 200 MeV per fission, calculate the number of fissions per second needed for a 500 megawatt reactor. (1MeV is 1.60 x 10^-13 Joules ) Assume (unrealistically) an efficiency of 100%.

1) 1.50 x 10²⁴

2) 1.56 x 10¹⁹

3) 1.45 x 10²²

4) 2.50 x 10⁶

Option 1

Option 2 Solution:

Option 3

Option 4

6 Atoms and Nuclei

When an electron and a position collide:

(1) The mass is not converted into energy
(2) They Repel each other
(3) The total mass is converted into energy
(4) Part of the mass is converted into energy

Option 1

Option 2

Option 3 Solution:

Option 4

7 Properties of Bulk Matter

Two soap bubbles A and B are kept in a closed chamber where the air is maintained at pressure 8N/m². The radii of bubbles A and B are 2 cm and 4 cm, respectively. Surface tension of the soap-water used to make bubbles is 0.04 N/m. Find the ratio nB/nA, where nA and nB are the number of moles of air in bubbles Aand B, respectively. [Neglect the effect of gravity.]

1) 2
2) 4
3) 6
4) 8

Option 1

Option 2

Option 3 Solution:

Solution

Excess pressure for soap bubble, $$\Delta P=\dfrac{4T}{R}$$ where $$T=$$ surface tension.
Here, $$\displaystyle P_A=P_0+\frac{4T}{R_A}=8+\frac{4(0.04)}{0.02}=16 N/m^2$$
and $$\displaystyle P_B=P_0+\frac{4T}{R_B}=8+\frac{4(0.04)}{0.04}=12 N/m^2$$
As $$PV=nRt , n\propto PV$$
Thus, $$\displaystyle \frac{n_A}{n_B}=\frac{P_AV_A}{P_BV_B}=\frac{16\times (4/3)\pi(0.02)^3}{12\times (4/3)\pi(0.04)^3}=1/6$$
$$\displaystyle \therefore \frac{n_B}{n_A}=6$$

Option 4

8 Gravitation

At what height from the surface of earth thegravitation potential and the value of g are –5.4 × 10⁷ J kg^–2 and 6.0 ms^–2 respectively ?Take the radius of earth as 6400 km :

1) 2600 km
2) 1600 km
3) 1400 km
4) 2000 km

Option 1 Solution:

Solution

r= 0.9×10⁷ m =9000 Km
Now height from surface (h) = height (r) - radius (R)
Height = 8000 – 6400 =2600 km

Option 2

Option 3

Option 4

9 Alternating Currents

Find the value of current through an inductance of $$0.7H$$ and negligible resistance when connected to an ac source of $$220V$$ and frequency $$50Hz$$. Take $$\pi=\cfrac{22}{7}$$.

1) 1 A
2) 4 A
3) 8 A
4) 2 A

Option 1 Solution:

Solution

For pure inductive circuit, Current lag by $$\dfrac { \pi }{ 2 } \\ $$

inductive reactance $${ X }_{ L }=\omega L=2\pi fL\\ { X }_{ L }=2\times \dfrac { 22 }{ 7 } \times 0.7\times 50\\ { X }_{ L }=220$$

$$ Current \ through \ inductor \ i=\dfrac { V }{ { X }_{ L } } =\dfrac { 220 }{ 220 } =1Amps$$

Option 2

Option 3

Option 4

10 Thermodynamics

A metal rod expands by 1.00 mm when it is heated from 20 C to 80 C. Another rod of the same material, twice as long as the first at 20 C is heated from 20C to 50 C. The second rod will expand by

(1) 0.25 mm
(2) 0.50 mm
(3) 1.00 mm
(4) 2.00 mm

Option 1

Option 2 Solution:

Option 3

Option 4

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