CUET · PHYSICS · PYQ PAPER 2025
A current of 15 A flows in a wire of cross-sectional area \( 5 \text{ mm}^2 \) with a drift velocity of \( 3 \times 10^{-3} \text{ m/s} \). The electron density in the wire will be :
- A \( 489 \times 10^{25} \text{ m}^{-3} \)
- B \( 546 \times 10^{25} \text{ m}^{-3} \)
- C \( 725 \times 10^{23} \text{ m}^{-3} \)
- D \( 625 \times 10^{25} \text{ m}^{-3} \)
Answer & Solution
Correct Answer
(D) \( 625 \times 10^{25} \text{ m}^{-3} \)
Step-by-step Solution
Detailed explanation
\( n = \frac{I}{e A v_d} \) \( n = \frac{15}{(1.6 \times 10^{-19})(5 \times 10^{-6})(3 \times 10^{-3})} \) \( n = \frac{15}{24 \times 10^{-28}} \) \( n = 0.625 \times 10^{28} \) \( n = 625 \times 10^{25} \text{ m}^{-3} \)
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\(KMnO _4\) is prepared by the fusion of \(MnO _2\) with an alkali metal hydroxide and an oxidizing agent like \(KNO _3\) to give a dark-green manganate ion which disproportionate to give permanganate as follows.
\(\begin{array}{l}2 MnO _2+4 KOH + O _2 \rightarrow 2 K_2 MnO _4+2 H _2 O \\ 3 K_2 MnO _4+4 H ^{+} \rightarrow 2 KMnO _4+ MnO _2+2 H _2 O \end{array}\)
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