Download App

STD 11 - 5. work,energy,power and collision — NEET STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceNEETSTD 11 - 5. work,energy,power and collision4 Marks
MCQ
If ${W_1},\,{W_2}$ and ${W_3}$ represent the work done in moving a particle from $A$ to $B$ along three different paths $1, 2$ and $3$ respectively (as shown) in the gravitational field of a point mass m, find the correct relation between ${W_1},\,{W_2}$ and ${W_3}$
  • A
    ${W_1} > {W_2} > {W_3}$
  • ✓
    ${W_1} = {W_2} = {W_3}$
  • C
    ${W_1} < {W_2} < {W_3}$
  • D
    ${W_2} > {W_1} > {W_3}$
✓

Answer

Correct option: B.
${W_1} = {W_2} = {W_3}$
b
(b)Gravitational force is a conservative force and work done against it is a point function i.e. does not depend on the path.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

NEET STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

A long straight wire along the $z$-axis carries a current $I$ in the negative $z$ direction. The magnetic vector field $\mathop B\limits^ \to $ at a point having coordinates $(x, y)$ in the $z = 0$ plane is
→
Consider two $npn$ transistors as shown in figure. If $0\,Volts$ corresponds to false and $5\, Volts$ correspond to true then the output at $C$ corresponds to
→
A bead is moving in medium $1$ with a uniform speed of $1\, m \,s^{-1}$ for $2.5\, s$. Then it enters into air and falls freely under gravity for $2\, m$. Finally, it enters medium $2$ and immediately moves with uniform speed for $1.5\, s$. The total distance the bead has traveled, is.........$m$ $(g = 10\, m/s^2$):
→
Light rays from a source are incident on a glass prism of index of refraction $\mu $ and angle of prism $\alpha $. At near normal incidence, the angle of deviation of the emerging rays is
→
A ray of light is incident normally on one of the face of a prism of angle $30^o$ and refractive index $\sqrt 2 $. The angle of deviation will be.....$^o$
→
The temperature of argon, kept in a vessel, is raised by $1^\circ C$ at a constant volume. The total heat supplied to the gas is a combination of translational and rotational energies. Their respective shares are
→
Two long current carrying thin wires, both with current $I$, are held by insulating threads oflength $L$ and are in equilibrium as shown in the figure, with threads making an angle '$\theta$' with the vertical. If wires have mass $\lambda$ per unit length then the value of $l$ is 
($g =$ gravitational acceleration)
→
The instantaneous voltages at three terminals marked $\mathrm{X}, \mathrm{Y}$ and $\mathrm{Z}$ are given by

$V_x=V_0 \sin \omega t$ $V_y=V_0 \sin \left(\omega t+\frac{2 \pi}{3}\right) \text { and }$ $V_z=V_0 \sin \left(\omega t+\frac{4 \pi}{3}\right)$

An ideal voltmeter is configured to read rms value of the potential difference between its terminals. It is connected between points $\mathrm{X}$ and $\mathrm{Y}$ and then between $\mathrm{Y}$ and $\mathrm{Z}$. The reading(s) of the voltmeter will be

$[A]$ $\mathrm{V}_{\mathrm{XY}}^{\mathrm{mms}}=\mathrm{V}_0 \sqrt{\frac{3}{2}}$

$[B]$ $\mathrm{V}_{\mathrm{YZ}}^{\mathrm{mms}}=\mathrm{V}_0 \sqrt{\frac{1}{2}}$

$[C]$ $\mathrm{V}_{\mathrm{XY}}^{\mathrm{mms}}=\mathrm{V}_0$

$[D]$ independent of the choice of the two terminals

→
A hydrogen atom makes a transition from $n = 2$ to  $n = 1$ and emits a photon. This photon strikes a doubly ionized lithium atom $(z = 3)$  in excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for the process is
→
One mole of an ideal gas undergoes two different cyclic processes I and II, as shown in the $P-V$ diagrams below. In cycle I, processes $a, b, c$ and $d$ are isobaric, isothermal, isobaric and isochoric, respectively. In cycle II, processes $a^{\prime}, b^{\prime}, c^{\prime}$ and $d^{\prime}$ are isothermal, isochoric, isobaric and isochoric, respectively. The total work done during cycle I is $W_I$ and that during cycle II is $W_{I I}$. The ratio $W_I / W_{I I}$ is . . . .

(image)

→