A ladder 5 m in length is resting against vertical wall. The bottom of the ladder is pulled along the ground a

Miscellaneous Exercise 2(I) · Q102

Q. A ladder 5 m in length is resting against vertical wall. The bottom of the ladder is pulled along the ground away from the wall at the rate of 1.5 m/sec. The length of the higher point of ladder when the foot of the ladder is 4.0 m away from the wall decreases at the rate of (A) 1 (B) 2 (C) 2.5 (D) 3

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✓ Free question

$x^2+y^2=25$ (ladder length fixed at $5$ ). Differentiating: $2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0 \Rightarrow \dfrac{dy}{dt}=-\dfrac{x}{y}\dfrac{dx}{dt}$ .

At $x=4$ : $y=\sqrt{25-16}=\sqrt9=3$ .

$\dfrac{dy}{dt}=-\dfrac{4}{3}(1.5)=-2$ m/sec.

The negative sign shows the height is decreasing, at $2$ m/sec.

✓Final answer

(B) 2

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set SEM31 markMCQ
$10\ \text{cm}^3/\text{sec}$
(a) $1{\cdot}5\ \text{cm}^2/\text{sec}$
(b) $1{\cdot}8\ \text{cm}^2/\text{sec}$
(c) $2\ \text{cm}^2/\text{sec}$
(d) $1{\cdot}25\ \text{cm}^2/\text{sec}$
›Reveal solutionSolution

Relate $\tfrac{dS}{dt}$ to $\tfrac{dV}{dt}$ through $\tfrac{dr}{dt}$ ; the surface area changes at $\tfrac{20}{r}=1{\cdot}25\ \text{cm}^2/\text{sec}$ .

This is a classic related-rates problem from the NCERT/CBSE Class 12 application of derivatives chapter.

Volume $V=\tfrac43\pi r^3$ , so

$\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt} = 10 \Rightarrow \frac{dr}{dt} = \frac{10}{4\pi r^2}.$

Surface area $S = 4\pi r^2$ , so

$\frac{dS}{dt} = 8\pi r\frac{dr}{dt} = 8\pi r\cdot\frac{10}{4\pi r^2} = \frac{20}{r}.$

At $r=16$ cm:

$\frac{dS}{dt} = \frac{20}{16} = 1{\cdot}25\ \text{cm}^2/\text{sec}.$

✓Final answer
$= 1{\cdot}25\ \text{cm}^2/\text{sec}$
CBSE 2019Set ANNUAL1 markMCQ
$Q. The rate of increase of a side of a square is 1 cm/sec. The rate of increase of area of the square, when length of a side of the square is 2 cm, is$
(a) 4 cm^2/sec
(b) 8 cm^2/sec
(c) 1 cm^2/sec
(d) 2 cm^2/sec
›Reveal solutionSolution

Relate $dA/dt$ to $ds/dt$ via $A=s^2$ , then substitute $s=2$ .

Let $s$ be the side length, $A=s^2$ the area. Given $\dfrac{ds}{dt}=1$ cm/sec.

Differentiate $A=s^2$ w.r.t. time:

$\dfrac{dA}{dt}=2s\dfrac{ds}{dt}$

At $s=2$ cm:

$\dfrac{dA}{dt}=2(2)(1)=4 \text{ cm}^2/\text{sec}$

✓Final answer
$4\text{ cm}^2/\text{sec}$

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