Units And Measurement And Dimensions JEE Main Physics: Authentic Shift PYQs (2021-2026) with 2.5-Min Shortcut Tricks & Trap Defense

Correct Answer: (d)

Explanation: To find the density, we use the formula: $\rho = \frac{m}{V}$ where • $ m = 49.53 \, \text{g} $ • $ V = 1.5 \, \text{cm}^3 $ Step 1: Calculate the density $\rho = \frac{49.53 \, \text{g}}{1.5 \, \text{cm}^3} \approx 33.02 \, \text{g/cm}^3$ Step 2: Consider Significant Figures The mass $49.53\, \text{g}$ has 4 significant figures. The volume $1.5\, \text{cm}^3$ has 2 significant figures. When dividing, the final answer must have the same number of significant figures as the value with the fewest significant figures. Here, that is 2 significant figures. Step 3: Round to 2 Significant Figures Rounding $33.02$ to 2 significant figures gives: $33 \, \text{g/cm}^3$ Among the options provided, the value that closely represents this magnitude is: Option D: 33.0 Thus, the correct answer is Option D.

Correct Answer: (c)

Explanation: Given, $S=\frac{\sqrt{a b}}{c^3 d^4}$ Percentage error in $S$ is given as, $\frac{\Delta s}{s} \times 100=\frac{1}{2} \times \frac{\Delta a}{a} \times 100+\frac{1}{2} \times \frac{\Delta b}{b} \times 100$ $ \begin{aligned} &\begin{aligned} & +3 \times \frac{\Delta c}{c} \times 100+4 \times \frac{\Delta d}{d} \times 100 \\ = & \frac{1}{2} \times 2 \%+\frac{1}{2} \times 1 \%+3 \times 1 \%+4 \times 1 \% \\ = & 1 \%+\frac{1}{2} \%+3 \%+4 \% \\ = & 8.5 \% \end{aligned}\\ &\text { Which is nearest to } 9 \% \text {. } \end{aligned} $

Correct Answer: (b)

Explanation: Given, $ T = 2\pi \sqrt{\frac{l}{g}} $ To find the error in $g$, first write the formula in terms of $g$: $ g = \frac{4\pi^2 l}{T^2} $ For percentage error, constants like $4\pi^2$ do not contribute any error. Since $l$ is in numerator and $T^2$ is in denominator, $ \frac{\Delta g}{g} \times 100 = \frac{\Delta l}{l} \times 100 + 2\frac{\Delta T}{T} \times 100 $ Now, length is about $ l = 80 \text{ cm} $ and its accuracy is $ \Delta l = 0.1 \text{ cm} $ So, $ \frac{\Delta l}{l} \times 100 = \frac{0.1}{80} \times 100 = 0.125\% $ The period is about $1.5$ s. Time for $50$ oscillations is $ t = 50 \times 1.5 = 75 \text{ s} $ The least count of stopwatch is $0.1$ s, so $ \Delta t = 0.1 \text{ s} $ Since $ T = \frac{t}{50} $ the fractional error in $T$ is same as the fractional error in $t$: $ \frac{\Delta T}{T} = \frac{\Delta t}{t} = \frac{0.1}{75} $ So, $ 2\frac{\Delta T}{T} \times 100 = 2 \times \frac{0.1}{75} \times 100 = 0.267\% $ Hence, $ \frac{\Delta g}{g} \times 100 = 0.125 + 0.267 = 0.392\% $ Nearly, $ \boxed{0.4\%} $ So the correct option is **Option B: $0.4$**.

Correct Answer: (a)

Explanation: The dimensions of force $[\mathrm{F}]=\left[M L \mathrm{~T}^{-2}\right]$ and density $[d]=\left[\mathrm{ML}^{-3} \mathrm{~T}^0\right]$ From the given relation, $F=\frac{y}{\sqrt{d}}$ $\Rightarrow \quad y=F \sqrt{d}$ Substituting the above dimensions, we get $\begin{aligned} {[y] } & =[\mathrm{F}][\mathrm{d}]^{1 / 2}=\left[M L \mathrm{~T}^{-2}\right]\left[\mathrm{ML}^{-3} \mathrm{~T}^0\right]^{1 / 2} \\ & =\left[\mathrm{M}^{3 / 2} \mathrm{~L}^{-1 / 2} \mathrm{~T}^{-2}\right] \end{aligned}$

Correct Answer: (c)

Explanation: The dimensions of Planck's constant, $[\mathrm{h}]=\frac{[\mathrm{E}]}{[\mathrm{v}]}=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^{-1}\right]}=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]$ and that moment of inertia, $\begin{aligned} & {[I]=[M][R]^2=[M][L]^2=\left[\mathrm{ML}^2 \mathrm{~T}^0\right]} \\ \therefore\quad & \frac{[\mathrm{h}]}{[I]}=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]}{\left[\mathrm{ML}^2 \mathrm{~T}^0\right]}=\left[\mathrm{T}^{-1}\right]=[\nu] \end{aligned}$ Thus, the ratio of dimensions of Planck's constant to that at moment of inertia is the dimensions of frequency.