A carpet of mass M made of a material is roll | vedclass

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

Question

(B) The density $\rho = M/V$ of the carpet remains constant.Let $M_1 = M$ and $R_1 = R$. The volume $V = \pi R^2 l$,where $l$ is the length of the car

Vedclass · Accepted Answer

$\frac{7}{8} MgR$

Vedclass · Answer

(B) The density $ho = M/V$ of the carpet remains constant.Let $M_1 = M$ and $R_1 = R$. The volume $V = \pi R^2 l$,where $l$ is the length of the carpet.When unrolled to radius $R_2 = R/2$,the mass $M_2$ of the carpet remaining in the rolled form is proportional to the volume of the cylinder.$M_2 = \frac{M}{\pi R^2 l} 	imes \pi (R/2)^2 l = \frac{M}{4}$.The potential energy of the initial rolled carpet (assuming center of mass at height $R$) is $U_1 = MgR$.The potential energy of the final rolled carpet (with radius $R_2 = R/2$ and mass $M_2 = M/4$) is $U_2 = M_2 g R_2 = (M/4) g (R/2) = \frac{1}{8} MgR$.The change in potential energy is $\Delta U = U_1 - U_2 = MgR - \frac{1}{8} MgR = \frac{7}{8} MgR$.

$A$ carpet of mass $M$ made of a material is rolled along its length in the form of a cylinder of radius $R$ and kept on a rough floor. If the carpet is unrolled without sliding to a radius $R/2$,the change in potential energy is ($g=$ acceleration due to gravity).

Similar Questions

$A$ body of mass $2\, kg$ is situated at a height $5\, m$ above the ground. Its potential energy is reported as $70\, J$. The potential energy of a $3\, kg$ body situated at a height $1\, m$ above the ground will be reported as ................ $J$.

The work done in pumping out water from a cubical vessel of height $1 \ m$ is approximately ........ $J$ (Take $g = 10 \ m/s^2$ and density of water $\rho = 1000 \ kg/m^3$).

Potential energy as a function of $r$ is given by $U = \frac{A}{r^{10}} - \frac{B}{r^{5}}$,where $r$ is the interatomic distance,and $A$ and $B$ are positive constants. The equilibrium distance between the two atoms will be

The potential energy of a diatomic molecule is given by $U = \frac{A}{r^{12}} - \frac{B}{r^6}$,where $A$ and $B$ are positive constants. The distance $r$ between them at equilibrium is:

$A$ particle is released from rest at the point $x = a$ and moves along the $x$ axis subject to the potential energy function $U(x)$ shown. The particle:

Vedclass Test Series

Exam Paper Generator

Online Exam Module