A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration (a) is varying with time t as a = k2 rt2, where k is a constant. The power delivered to the particle by the force acting on it is given as

mk2 r2 t2

mk2 r2 t2

Ask & Answer Question

Physics Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Five Laws of Motion

A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration (a) is varying with time t as a = k² rt², where k is a constant. The power delivered to the particle by the force acting on it is given as

Option 1 -

zero

Option 2 -

mk² r² t²

Option 3 -

mk² r² t²

Option 4 -

mk² rt

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

2 months ago

Correct Option - 3

Detailed Solution:

a = k²rt²

$\Rightarrow \frac{v^{2}}{r} = k^{2} r t^{2}$

$\Rightarrow v = k r t$

$a_{t} = \frac{d v}{d t} = k r$

⇒ tangential force, F_t = ma_t = mkr

$∴ P o w e r d e l i v e r e d, P = F_{t} v = (m k r) (k r t) = m k^{2} r^{2} t$

Note → Power delivered by centripetal force will be zero.

a = k2rt2 <math> <mrow> <mo>⇒</mo> <mfrac> <mrow> <msup> <mrow> <mi>v</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mi>r</mi> </mrow> </mfrac> <mo>=</mo> <msup> <mrow> <mi>k</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mi>r</mi> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math>                         <div><div><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1752219354phpnlljQM.jpeg" alt="" width="101" height="81" loading="lazy"></picture>           <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1813744982"> </o:OLEObject></xml><! [endif]-> </div></div> <math> <mrow> <mo>⇒</mo> <mi>v</mi> <mo>=</mo> <mi>k</mi> <mi>r</mi> <mi>t</mi> </mrow> </math> <math> <mrow> <msub> <mrow> <mi>a</mi> </mrow> <mrow> <mi>t</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mi>d</mi> <mi>v</mi> </mrow> <mrow> <mi>d</mi> <mi>t</mi> </mrow> </mfrac> <mo>=</mo> <mi>k</mi> <mi>r</mi> </mrow> </math>              <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1813744983"> </o:OLEObject></xml><! [endif]-> ⇒ tangential force, Ft = mat = mkr <math> <mrow> <mo>∴</mo> <mi>P</mi> <mi>o</mi> <mi>w</mi> <mi>e</mi> <mi>r</mi> <mtext> </mtext> <mtext> </mtext> <mi>d</mi> <mi>e</mi> <mi>l</mi> <mi>i</mi> <mi>v</mi> <mi>e</mi> <mi>r</mi> <mi>e</mi> <mi>d</mi> <mo>, </mo> <mi>P</mi> <mo>=</mo> <msub> <mrow> <mi>F</mi> </mrow> <mrow> <mi>t</mi> </mrow> </msub> <mi>v</mi> <mo>=</mo> <mrow> <mo> (</mo> <mrow> <mi>m</mi> <mi>k</mi> <mi>r</mi> </mrow> <mo>)</mo> </mrow> <mrow> <mo> (</mo> <mrow> <mi>k</mi> <mi>r</mi> <mi>t</mi> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mi>m</mi> <msup> <mrow> <mi>k</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <msup> <mrow> <mi>r</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mi>t</mi> </mrow> </math>              <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1028" DrawAspect="Content" ObjectID="_1813744984"> </o:OLEObject></xml><! [endif]-> Note → Power delivered by centripetal force will be zero.

0 Upvotes 0 Downvotes

Similar Questions for you

A travelling microscope is used to determine the refractive index of a glass slab. If 40 divisions are there in 1cm on main scale and 50 Vernier scale divisions are equal to 49 main scale division, then least count of the travelling microscope is __________× 10^-6 m.

Vishal Baghel

Let ‘x’ be the value of one division on main scale

$x = \frac{1}{40} c m$

Let ‘y’ be the value of one division on vernier

Now given

50y = 49 x

$y = \frac{49 x}{50}$

$\begin{array}{l} = 5 \times 1 0^{- 4} c m \\ = 5 \times 1 0^{- 6} m \\ 5 \end{array}$

A ball of mass 0.5 kg is dropped from the height of 10m. The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is __________m.

[Use g = 10m/s²]

Payal Gupta

Let ‘h’ be the height at which velocity becomes equal to magnitude of Acceleration

v = g = 10

v = u + at

10 = 0 + 10t

t = 1 sec

$h = u t + \frac{1}{2} a t^{2}$

$= 0 \times 1 + \frac{1}{2} \times 10 \times 1 \times 1$

= 5m

In the arrangement shown in figure a₁, a₂, a₃ and a₄ are the accelerations of masses m₁, m₂, m₃ and m₄ respectively. Which of the following relation is true for this arrangement?

Vishal Baghel

$\sum_{}^{} \vec{T} . \vec{a} = 0$

$\frac{T}{2} a_{1} + \frac{T}{4} a_{2} + \frac{T}{8} a_{3} + \frac{T}{8} a_{4} = 0$

$\Rightarrow 4 a_{1} + 2 a_{2} + a_{3} + a_{4} = 0$

A thin circular ring of mass M and radius R is rotating with a constant angular velocity 2 rads^-1 in a horizontal plane about an axis vertical to its plane and passing through the centre of the ring. If two objects each of mass m be attached gently to the opposite ends of a diameter of ring, the will then rotate with an angular velocity (in rads^-1).

Payal Gupta

By conservation of Angular momentum

L_i = L_f

$M R^{2} ω = (m R^{2} + 2 m R^{2}) ω'$

$\frac{2 M}{M + 2 m} = ω'$

$\Rightarrow$ $ω' = \frac{2 M}{M + 2 m}$

A block of metal weighing 2kg is resting on a frictionless plane (as shown in figure). It is struck by a jet relasing water at a rate of 1 kgs^-1 and at a speed of 10 ms^-1. Then, the initial acceleration of the block, in ms^-2, will be: