Physics · 2025
JEE Main · 29 January 2025, Shift 2 · Q46
A physical quantity Q is related to four observables a, b, c, d as follows: Q =(ab^4)/cd where, a =(60 ± 3) Pa; b =(20 ± 0.1) m; c =(40 ± 0.2) Nsm^-2…
A physical quantity Q is related to four observables $\displaystyle \mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ as follows :
$$\mathrm{Q}=\frac{\mathrm{ab}^4}{\mathrm{~cd}}
$$
where, $\displaystyle \mathrm{a}=(60 \pm 3) \mathrm{Pa} ; \mathrm{b}=(20 \pm 0.1) \mathrm{m} ; \mathrm{c}=(40 \pm 0.2) \mathrm{Nsm}^{-2}$ and $\displaystyle \mathrm{d}=(50 \pm 0.1) \mathrm{m}$, then the percentage error in Q is $\displaystyle \frac{x}{1000}$, where $\displaystyle x=$ $\displaystyle \_\_\_\_$.
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JEE Main 2025 Physics question, with the answer from NTA’s final answer key. Where our answers come from.