A bulb is placed at the centre of a circular | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $r = 10 \ m$ be the radius of the circular track.The speed of the man along the track is $v = r \frac{d	heta}{dt} = 10 \ m/sec$.Since $r = 10

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) Let $r = 10 \ m$ be the radius of the circular track.The speed of the man along the track is $v = r \frac{d	heta}{dt} = 10 \ m/sec$.Since $r = 10 \ m$, we have $10 \frac{d	heta}{dt} = 10$, which implies $\frac{d	heta}{dt} = 1 \ rad/sec$.Let $y$ be the position of the shadow on the wall at an angular distance $	heta$ from $P$.From the geometry, we have $	an 	heta = \frac{y}{r}$, so $y = r 	an 	heta$.Differentiating with respect to time $t$, we get $\frac{dy}{dt} = r \sec^2 	heta \cdot \frac{d	heta}{dt}$.At $	heta = 60^{\circ}$, we have $\sec(60^{\circ}) = 2$, so $\sec^2(60^{\circ}) = 4$.Substituting the values, $\frac{dy}{dt} = 10 	imes 4 	imes 1 = 40 \ m/sec$.Thus, the speed of the shadow is $40 \ m/sec$.

Similar Questions

The radius of a circle is increasing at a rate of $0.1 \text{ cm s}^{-1}$. Then the rate of change of area,when its radius is $5 \text{ cm}$,is .........

If $r$ is the radius of a spherical balloon at time $t$ and the surface area of the balloon changes at a constant rate $K,$ then ....

The displacement $x$ of a particle at time $t$ is given by $x = At^2 + Bt + C$, where $A, B, C$ are constants and $v$ is the velocity of the particle. Then the value of $4Ax - v^2$ is:

The rate of change of the volume of a sphere with respect to its surface area when the radius is $5 \ m$ is:

The surface area of a cube increases at a rate of $2 \ cm^2/sec$. The rate at which its volume increases when the length of its edge is $90 \ cm$ is ..... $cm^3/sec$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module