Physics JEE main pyQ (JEE/NEET PYQs)

Physics JEE main pyQ (JEE/NEET PYQs)

Practice fundamental concepts and calculations across core subjects.

Question 1: Maximum Fractional Error

A quantity $Q$ is formulated as $Q = X^{-2}Y^{\frac{3}{2}}Z^{-\frac{2}{5}}$. $X$, $Y$, and $Z$ are independent parameters which have fractional errors of $0.1$, $0.2$, and $0.5$, respectively in measurement. The maximum fractional error of $Q$ ($\frac{\Delta Q}{Q}$) is:

A) 0.6 B) 0.8 C) 0.7 D) 0.1

Solution 1: Error Propagation

The maximum fractional error is calculated by summing the absolute value of the powers multiplied by their respective fractional errors. All exponents are treated as positive to maximize the error.

$$Q = X^{-2}Y^{\frac{3}{2}}Z^{-\frac{2}{5}}$$ $$\frac{\Delta Q}{Q} = \left| -2 \right| \frac{\Delta X}{X} + \left| \frac{3}{2} \right| \frac{\Delta Y}{Y} + \left| -\frac{2}{5} \right| \frac{\Delta Z}{Z}$$ $$\frac{\Delta Q}{Q} = 2 \times (0.1) + 1.5 \times (0.2) + 0.4 \times (0.5)$$ $$\frac{\Delta Q}{Q} = 0.2 + 0.3 + 0.2$$ $$\frac{\Delta Q}{Q} = \mathbf{0.7}$$

**Correct Option: C**.

Question 2: Dielectric Constant from EM Wave Equation

Electric field of plane electromagnetic wave propagating through a non-magnetic medium is given by $E = 20\cos(2 \times 10^{10} t - 200x) \text{ V/m}$. The dielectric constant of the medium ($\epsilon_r$) is equal to: (Take $\mu_r = 1$)

(1) 9 (2) 2 (3) $\frac{1}{3}$ (4) 3

Solution 2: Wave Velocity and Dielectric Constant

1. **General Wave Equation:** Compare $E = 20\cos(2 \times 10^{10} t - 200x)$ with $E = E_0 \cos(\omega t - kx)$. $$\text{Angular frequency } (\omega) = 2 \times 10^{10} \text{ rad/s}$$ $$\text{Wave number } (k) = 200 \text{ rad/m}$$
2. **Wave Velocity in Medium ($v$):** $$v = \frac{\omega}{k} = \frac{2 \times 10^{10}}{200} = 1 \times 10^8 \text{ m/s}$$
3. **Relation with Dielectric Constant:** The wave speed in a medium is $v = \frac{c}{\sqrt{\mu_r \epsilon_r}}$. Given that the medium is non-magnetic ($\mu_r = 1$) and $c = 3 \times 10^8 \text{ m/s}$: $$v = \frac{c}{\sqrt{\epsilon_r}} \implies \sqrt{\epsilon_r} = \frac{c}{v}$$ $$\sqrt{\epsilon_r} = \frac{3 \times 10^8}{1 \times 10^8} = 3$$ $$\epsilon_r = 3^2 = \mathbf{9}$$

**Correct Option: (1)** (9).

Question 3: Apparent Depth and Refraction

A glass tumbler having inner depth of $17.5 \text{ cm}$ is kept on a table. A student starts pouring water ($\mu = 4/3$) into it while looking at the surface of water from the above. When he feels that the tumbler is half filled, he stops pouring water. Up to what height, the tumbler is actually filled?

(1) $11.7 \text{ cm}$ (2) $10 \text{ cm}$ (3) $7.5 \text{ cm}$ (4) $8.75 \text{ cm}$

Solution 3: Apparent Depth

The total depth of the tumbler is $D = 17.5 \text{ cm}$. When the student "feels" the tumbler is half-filled, it means the **apparent depth** of the water surface (as seen through the air) is half the total depth of the tumbler.

1. **Target Apparent Depth ($D'_{total}$):** $$D'_{total} = \frac{17.5 \text{ cm}}{2} = 8.75 \text{ cm}$$
2. **Apparent Depth Components:** This apparent depth is the sum of the apparent depth of the water ($D'_{water}$) and the apparent depth of the remaining empty space ($D'_{air}$), which is just the actual depth of the empty space. $$D'_{total} = D'_{water} + D'_{air} = 8.75 \text{ cm}$$
3. **Formulation:** Let $h$ be the actual height of the water (Actual Depth). The empty space is $17.5 - h$. $$\text{Apparent Depth of water } (D'_{water}) = \frac{\text{Actual Depth}}{\mu} = \frac{h}{4/3} = \frac{3h}{4}$$ $$\text{Apparent Depth of empty space } (D'_{air}) = 17.5 - h$$
4. **Calculation:** $$\frac{3h}{4} + (17.5 - h) = 8.75$$ $$17.5 - 8.75 = h - \frac{3h}{4}$$ $$8.75 = \frac{h}{4}$$ $$h = 8.75 \times 4 = \mathbf{35.0 \text{ cm}}$$

Wait! The actual depth cannot be greater than the total depth ($17.5 \text{ cm}$). This suggests the student meant the apparent depth of the **bottom of the tumbler** appears to be halfway up the tumbler.

1. **Target Apparent Height of Bottom ($D'_{bottom}$):** $$D'_{bottom} = \frac{17.5 \text{ cm}}{2} = 8.75 \text{ cm}$$
2. **Apparent Depth of Bottom:** The apparent depth of the bottom ($D'_{bottom}$) is the actual depth ($h$) divided by the refractive index ($\mu$). $$D'_{bottom} = \frac{h}{\mu}$$
3. **Calculation:** $$8.75 \text{ cm} = \frac{h}{4/3}$$ $$h = 8.75 \times \frac{4}{3} = \frac{35}{3} \approx 11.666 \text{ cm}$$

This still doesn't match the official answer of $10 \text{ cm}$ (Option 2). Let's re-read the setup: "When he feels that the tumbler is half filled, he stops pouring water."

This statement most likely means the student sees the **surface of the water** halfway down the tumbler from the top. Let $H=17.5 \text{ cm}$ be the total depth and $h$ be the actual height of water.

The student views the water surface from above. When the tumbler is half-filled, the distance from the top rim to the water surface is $\frac{H}{2} = 8.75 \text{ cm}$. This is the *actual* height of the empty space. Thus, the actual height of the water is also $h = \frac{H}{2} = \mathbf{8.75 \text{ cm}}$.

The question is tricky, as it uses the refraction concept ($\mu$) but the condition for stopping seems to be purely geometric (actual half-fill) or based on the apparent position of the bottom.

If we assume the question means that the **actual depth** is half of the total depth, then $h = 17.5/2 = \mathbf{8.75 \text{ cm}}$ (Option 4), but the official answer is (2) $10 \text{ cm}$.

Let's use the apparent depth formula again, but with the premise that the student is pouring until the bottom of the tumbler *appears* to be at the midpoint of the entire tumbler's depth. $$\text{Apparent height of bottom from top rim } = \text{Empty space} + \text{Apparent water depth}$$ $$8.75 = (17.5 - h) + \frac{h}{\mu}$$ $$8.75 = 17.5 - h + \frac{3h}{4}$$ $$h - \frac{3h}{4} = 17.5 - 8.75$$ $$\frac{h}{4} = 8.75 \implies h = 35 \text{ cm} \text{ (Impossible)}$$

Given the official answer of **$10 \text{ cm}$** ($\approx 11.7 \text{ cm}$ is not an option), the most plausible interpretation that yields a match to **$11.7 \text{ cm}$** (Option 1) is: **The apparent depth of the bottom is $11.7 \text{ cm}$** and the student stopped pouring when this value reached $8.75 \text{ cm}$.

Let's assume the question meant that the **apparent height of the bottom of the tumbler from the water surface** is $D' = 17.5/2 = 8.75 \text{ cm}$.

$$\text{Apparent depth of bottom } = \frac{h}{\mu} \implies 8.75 = \frac{h}{4/3} \implies h \approx 11.7 \text{ cm} \text{ (Option 1)}$$

**Conclusion based on NTA Answer (2):** The official answer is $10 \text{ cm}$. This value cannot be derived from standard apparent depth formulas based on the wording. However, in the context of competitive exams, if $10 \text{ cm}$ is the official answer, it might suggest a typo in the question's values or a highly unconventional interpretation. Since $11.7 \text{ cm}$ (Option 1) is mathematically derived from a common interpretation (apparent depth of water is half the total depth, $D' = D/2$), let's trust the logic over the potentially incorrect official answer key provided in the image.

**Re-analyzing the question for $10 \text{ cm}$:** If the total depth was $20 \text{ cm}$ and $h$ was $15 \text{ cm}$, the apparent depth would be $15/(4/3) = 11.25 \text{ cm}$. If the actual height was $10 \text{ cm}$, the apparent depth would be $10/(4/3) = 7.5 \text{ cm}$. The actual water height that gives $8.75 \text{ cm}$ apparent depth of the bottom is $11.7 \text{ cm}$.

**For consistency with standard physics principles, the result is $\mathbf{11.7 \text{ cm}}$ (Option 1).** Since the image explicitly states Option (2) is the NTA answer, we will note this discrepancy but provide the physically correct calculation for the most logical interpretation: $h \approx 11.7 \text{ cm}$.

Question 4: Half-life vs. Mean-life

The half life period of radioactive element $x$ ($T_{1/2}^x$) is same as the mean life time of another radioactive element $y$ ($\tau_y$). Initially they have the same number of atoms. Then:

(1) $x$ will decay faster than $y$. (2) $y$ will decay faster than $x$. (3) $x$ and $y$ have same decay rate initially and later on different decay rate. (4) $x$ and $y$ decay at the same rate always.

Solution 4: Radioactivity Comparison

The decay rate is proportional to the decay constant $\lambda$ (Rate $= \lambda N$). The decay is faster for the element with the larger $\lambda$.

1. **Relations:** $$\text{Half-life: } T_{1/2} = \frac{\ln(2)}{\lambda} = \frac{0.693}{\lambda}$$ $$\text{Mean-life: } \tau = \frac{1}{\lambda}$$
2. **Given Condition:** $T_{1/2}^x = \tau_y$ $$\frac{0.693}{\lambda_x} = \frac{1}{\lambda_y}$$
3. **Compare Decay Constants:** $$\lambda_x = 0.693 \lambda_y$$ Since $0.693 < 1$, it follows that $\lambda_x < \lambda_y$.

Since the decay constant of $y$ ($\lambda_y$) is greater than the decay constant of $x$ ($\lambda_x$), the element $\mathbf{y}$ decays faster than $\mathbf{x}$.

**Correct Option: (2)** ($y$ will decay faster than $x$).

Question 5: Capacitance in RC Charging Circuit

A capacitor is connected to a $20 \text{ V}$ battery through a resistance of $10\Omega$. It is found that the potential difference across the capacitor rises to $2 \text{ V}$ in $1 \mu\text{s}$. The capacitance of the capacitor is $\underline{\hspace{2em}} \mu\text{F}$.

Given: $\ln \left(\frac{10}{9}\right)=0.105$

(1) $9.52$ (2) $0.95$ (3) $0.105$ (4) $1.85$