Q

# Tell me? - Limit , continuity and differentiability - JEE Main-7

$2m$ ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate $25cm/sec$, then the rate $\left ( in\: \: cm/sec\right )$ at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is $1m$ above the ground is :

• Option 1)

$25$

• Option 2)

$\frac{25}{3}$

• Option 3)

$\frac{25}{\sqrt{3}}$

• Option 4)

$25\sqrt{3}$

Views

$x^{2}+y^{2}=4$                                   $\frac{\mathrm{d} y}{\mathrm{d} t}=-25$

$x\frac{\mathrm{d} x}{\mathrm{d} t}+y\frac{\mathrm{d} y}{\mathrm{d} t}=0$

$\sqrt{3}\frac{\mathrm{d} x}{\mathrm{d} t}\pm \times 25=0$

$\frac{\mathrm{d} x}{\mathrm{d} t}=\frac{25}{\sqrt{3}}\: \: \: cm/sec$

Option 1)

$25$

Option 2)

$\frac{25}{3}$

Option 3)

$\frac{25}{\sqrt{3}}$

Option 4)

$25\sqrt{3}$

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