CUET · MATHS · PYQ PAPER 2023

The volume of a cube is increasing at a rate of 9 cubic centimeter per second. Then rate of change of surface area when the length of an edge is 10 centimeter is:

A \(36.6 cm^2 / s\)
B \(0.36 cm^2 / s\)
C \(36 cm^2 / s\)
D \(3.6 cm^2 / s\)

Verified Solution

Answer & Solution

Correct Answer

(D) \(3.6 cm^2 / s\)

Step-by-step Solution

Detailed explanation

\(\frac{dV}{dt} = 3x^2 \frac{dx}{dt}\) \(9 = 3(10)^2 \frac{dx}{dt} \implies \frac{dx}{dt} = \frac{9}{300} = 0.03 \text{ cm/s}\) \(\frac{dA}{dt} = 12x \frac{dx}{dt}\) \(\frac{dA}{dt} = 12(10)(0.03) = 3.6 \text{ cm}^2/\text{s}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Match List-I with List-II

List-I (Differential Equations)	List-II (Order and degree)
(A) \(\frac{d y}{d x}+e^y=0\)	(I) order 2, degree not defined
(B) \(\frac{d^2 y}{d x^2}=\left[1+\left(\frac{d y}{d x}\right)^2\right]^{3 / 2]}\)	(II) order 2, degree 1
(C) \(\left(\frac{d^2 y}{d x^2}\right)^2+e^{\left(\frac{\text { 临 })}{}\right.}=0\)	(III) order 1, degree 1
(D) \(\frac{d^2 y}{d x^2}+x \frac{d y}{d x}-2 y=\log x ; x>0\)	(IV) order 2, degree 2

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From CUET

Based on the standard electrode potential given in the table

Half-cell reaction	(E./V) at 25.C)
\(F _2(g)+2 e ^{-} \rightarrow 2 F^{-}\)	\(2.87\)
\(NO _3^{-}+4 H ^{+}+3 e ^{-} \rightarrow NO ( g )+2 H _2 O\)	\(0.97\)
\(Cu ^{2+}+2 e ^{-} \rightarrow Cu ( s )\)	\(0.34\)
\(2 H ^{+}+2 e ^{-} \rightarrow H _2(g)\)	0.00
\(Zn ^{2+}+2 e ^{-} \rightarrow Zn ( s )\)	\(-0.76\)
\(L i ^{+}+ e ^{-} \rightarrow L i ( s )\)	\(-3.05\)

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