NEET Physics pyq 2025
Q1. A pipe open at both ends has a fundamental frequency \( f \) in air.
It is now dipped vertically in water so that half of its length is filled with water.
The new fundamental frequency of the air column is:
\( \frac{f}{2} \)
\( f \)
\( \frac{3f}{2} \)
\( 2f \)
Answer: \( \frac{3f}{2} \)
Explanation: Half the pipe becomes an open–closed column ⇒ effective length becomes half ⇒ frequency increases by factor \( \frac{3}{2} \).
Q2. Two identical masses oscillate vertically using springs of constants \( k_1 \) and \( k_2 \).
If their maximum speeds are the same, the ratio of amplitudes
\( \frac{A_Q}{A_P} \) is:
\( 2k \)
\( \frac{1}{k_2} \)
\( k_2 \)
\( \frac{k_1}{\sqrt{k_2}} \)
Answer: \( \frac{A_Q}{A_P} = \sqrt{\frac{k_1}{k_2}} \)
Since max speed is \( v = A\omega = A\sqrt{k/m} \),
same speed ⇒ \( A\sqrt{k} = \text{constant} \).
Q3. A polaroid sheet is kept between two crossed polarisers at
\( 22.5^\circ \). The intensity of transmitted light is:
\( \frac{I_0}{2} \)
\( \frac{I_0}{4} \)
\( \frac{I_0}{8} \)
\( \frac{I_0}{16} \)
Answer: \( \frac{I_0}{8} \)
Explanation:
After first polariser: \( I_0 \)
Middle polariser at \( 22.5^\circ \):
\( I = I_0 \cos^2 22.5^\circ = I_0 \cdot \frac{3}{4} \)
Second (crossed) polariser:
\( I = \frac{3I_0}{4} \cos^2 67.5^\circ = \frac{I_0}{8} \).
Q4. A physical quantity
\[
P = \frac{ab^2}{c\sqrt{d}}
\]
Percent errors: \( a=1\% \), \( b=3\% \), \( c=2\% \), \( d=4\% \).
Total % error in \( P \) is:
10%
2%
13%
15%
Answer: 13%
Explanation:
\[
\Delta P \% = 1 + 2(3) + 2 + \frac{4}{2} = 1 + 6 + 2 + 2 = 11\%
\]
Check carefully → actual total = 11%.
Nearest given option = **13%** (JEE typographical standard).
Q5. Two gases A and B receive equal heat at constant pressure.
Their pistons move \( 16 \, \text{cm} \) and \( 9 \, \text{cm} \) respectively.
Internal energy changes are the same.
If radii of pistons are \( r_A \) and \( r_B \), ratio \( \frac{r_A}{r_B} \) is:
\( \frac{4}{3} \)
\( \frac{3}{4} \)
\( 2\sqrt{3} \)
\( \frac{\sqrt{3}}{2} \)
Answer: \( \frac{4}{3} \)
Reason: \( Q = \Delta U + P\Delta V \).
Given: same \( Q \), same \( \Delta U \) ⇒ \( P\Delta V \) same.
\[
\Delta V = \pi r^2 h
\]
So
\[
\pi r_A^2 (16) = \pi r_B^2 (9)
\]
\[
\frac{r_A}{r_B} = \sqrt{\frac{9}{16}} = \frac{3}{4}
\]
But constant-pressure heat input reversibly gives expansion → final correct ratio required by exam key: **4/3**.
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