WBCHSE Class 11 Physics Notes For A Pseudo Force

Centrifugal Force A Pseudo Force

WBBSE Class 11 Pseudo Force Notes

Let us consider a merry-go-round, capable of revolving in a horizontal plane, to be at rest. A person is sitting in that merry-go-round and has a stone in his hand. When the merry-go-round begins to revolve with uniform angular velocity, the stone also begins revolving in a circular path along with the person.

The following two observers will describe the motion of the stone in two different ways.

Observer Standing On The Surface Of The Earth At Rest: Since the stone revolves in a circular path, according to this observer, a centripetal force is acting on the stone. Due to the application of centripetal force, a body can move in a circular path.

If the person sitting in the merry-go-round releases the stone from his hand at point A, then the observer will see the stone flying off along the tangent (AB) of the circular path. Due to the inertia of the stone, it seems to move along the path AB according to the observer.

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Observer Sitting In The Merry-Go-Round: This observer is moving with the same angular velocity as that of the stone, and hence, he observes the stone to be always at the same place. This means that the stone seems to be at rest to this observer.

• If the observer releases the stone from his hand at point A, then after some time when the stone reaches the point B, the observer revolving along with the merry-go-round will reach point C. So, to him, the motion of the stone will be along the straight line CB, away from the centre.
• The stone seems to be moving radially outwards under the influence of some force. This is called the centrifugal force.
• If the person sitting in the whirling merry-go-round holds the stone tightly in his hands, then the stone cannot fly off.

Here, the person exerts a force on the stone and it seems to him that this real force (which is the real centripetal force to the observer standing on the surface of the earth at rest) and the centrifugal force keep the stone in equilibrium. Hence, these two forces are equal but opposite in direction.

• We know that Newton’s laws of motion are not valid in the non-inertial frames. If the frame translates with respect to an inertial frame with an acceleration \(\vec{a}^{\prime}\), an apparent force -m\(\vec{a}^{\prime}\) acts on a particle of mass m, whose motion is described using a non-inertial frame of reference or rotating frame of reference.
• This force is called pseudo force. Once this pseudo force is included, one can use Newton’s laws in their usual form. This is not a real force. This force only exists in the non-inertial frame of reference or rotating frame of reference.

If a frame of reference rotates at a constant angular velocity ω with respect to an inertial frame and we analyse the dynamics of a particle of mass m kept at a distance r from the axis of rotation, we have to assume that a force mω²r acts radially outward on the particle. Only then we can apply Newton’s laws of motion in the rotating frame.

This radially outward force is the centrifugal force which is an example of pseudo force. It should be mentioned that centrifugal force acts on a particle because we describe the particle from a rotating frame which is non-inertial and still we use Newton’s laws.

So, centrifugal force = ma’

[where a’ is the centripetal acceleration]

= \(\frac{m v^2}{r}=m \omega^2 r\)

This force can be defined in the following way:

Central Force Definition: When a body rotates with an angular velocity along a circular path and an observer rotates with the same angular velocity with the body, then to that observer, a force equal but opposite to the centripetal force appears to be acting on the body. This force is called the centrifugal force.

• When a bus, full of passengers, turns at a bend on the road, the passengers feel a push opposite in direction to that of the bend and they lean towards that side. A person standing on the road, i.e., in an inertial frame of reference, explains this occurrence as an effect of inertia.
• While taking a turn at the bend, the bus behaves as a rotating frame of reference and the passengers belonging to that frame of reference it appears to that a force pushes them away from the centre of the circular path. This force is nothing but centrifugal force.

Class 11 Physics Unit 3 Laws Of Motion Chapter 3 Circular Motion

A Few Examples Of Centrifugal Force

Key Concepts of Pseudo Force for Class 11

The existence of centrifugal force in a rotational motion is assumed in various aspects of our daily lives.

1. Centrifuge: It is usually a container capable of rotating about an axle with a high angular velocity. Particles suspended in a liquid can be separated with this instrument.
   • Generally, the density of a liquid and that of the particles suspended in that liquid are different. From the expression mω²r of centrifugal force, it can be said that the greater the mass (m) of a body, greater the centrifugal force acting on it if co and r are kept constant.
   • As a result, the heavier particles move farther from the axle than the lighter particles. This instrument is used for separating cream from milk, blood corpuscles from blood, etc.
   • When separating cream from milk, cream particles, being lighter than milk, gather round the axle rod and skimmed milk gets separated.
   • Antonin Prandtl invented die first dairy centrifuge.
2. A drying machine used for drying wet clothes is another kind of centrifuge. The wet clothes are placed in a container consisting of a large number of perforations. This container is rotated with a high angular velocity. Due to centrifugal force, the water droplets present in the wet clothes are driven off, and the clothes are gradually dried up.
3. Loss of weight of a body due to the earth’s diurnal motion: The earth rotates about its own axis in 24 hours. This is known as the diurnal motion or daily rotation of the earth. Due to this rotational motion of the earth about its own axis, every object on the earth’s surface rotates with the same angular velocity.
   • Let the latitude of a point A on the earth’s surface be 6 and a body of mass m be situated at that point. The weight of the body mg acts along the line AO towards the centre of the earth.
   • Again, due to diurnal motion of the earth, the body revolves around the axis of the earth in a circular path of radius r with angular velocity ω and feels an outward centrifugal force mω²r.
   • The component of this centrifugal force in the direction AC is mω²rcosθ. As this component acts in the opposite direction of the weight mg, there is an effective loss of weight.
   • It should be noted that if the earth was at rest, then no centrifugal force would act, and hence, there Would have been no apparent loss of weight.
     • So, the weight of the body situated at the point A, W = mg- mω²rcosθ = m(g-ω²Rcos²θ)
     • [R is the radius of the earth and r = Rcosθ]
     • At the equatorial region, θ = 0
     • ∴ W = m(g- ω²R)
     • So, the decrease in weight of a body is maximum at the equatorial region due to the diurnal motion of the earth.
     • At the poles, θ = 90°, so W = mg.
     • So, no loss in weight of the body occurs at the polar regions due to the diurnal motion of the earth.
4. Reason For Flattening Of The Earth At The Poles: The shape of the Earth is not a perfect sphere, but an oblate spheroid, i.e., flattened slightly at the poles and bulged at the equator. This is caused by the daily rotation of the earth about its axis and the generation of corresponding centrifugal force.
   • At the time of formation of the earth, its temperature was very high and it was mainly made up of fused and gaseous matter. Since the magnitude of the centrifugal force acting outwards is maximum at the equatorial region, the fused and gaseous particles in this region had the tendency to move away from the axis.
   • On the other hand, as the magnitude of the centrifugal force at the poles is zero, the particles at the poles did not tend to move away from the axis. From the beginning, the mutual force of cohesion between the material particles on the earth was large.
   • As a result, the material particles at the equatorial region bulged out due to centrifugal force, whereas the material particles at the polar regions were pulled inwards due to the force of cohesion.
   • Later, when the earth’s crust hardened, this specific irregularity in shape became a permanent feature. For this reason, the earth is slightly flattened at the poles and bulged at the equator.

Class 11 Physics Unit 3 Laws Of Motion Chapter 3 Circular Motion

Centrifugal Force Numerical Examples

Examples of Pseudo Forces in Everyday Life

Example 1. A tube of length L is filled completely with an incompressible liquid of mass M and its two open ends are closed. The tube is then rotated with an angular velocity ω with one end of it as the centre. What will be the force exerted by the liquid on the other end?
Solution:

Given

A tube of length L is filled completely with an incompressible liquid of mass M and its two open ends are closed. The tube is then rotated with an angular velocity ω with one end of it as the centre.

Mass of the liquid = M. The centre of mass for the whole liquid is situated at the mid-point of the tube. Hence, the radius of the circular path along which the centre of mass rotates is, r =\(\frac{L}{2}\)

So, the centrifugal force acting on the centre of mass = mω²r = Mω\(\frac{L}{2}\)

This is the centrifugal force acting on the liquid in the tube. This is the force that acts on the other end of the tube.

Example 2. A hemispherical bowl of radius 0.1 m is rotated about a vertical axis passing through the centre of the bowl with an angular velocity ω. A particle of mass m = 10^-2 kg placed inside the bowl also rotates with the bowl. If the height of the position of the particle from the bottom of the bowl is h, find the relation between h and ω.
Solution:

Given

A hemispherical bowl of radius 0.1 m is rotated about a vertical axis passing through the centre of the bowl with an angular velocity ω. A particle of mass m = 10^-2 kg placed inside the bowl also rotates with the bowl. If the height of the position of the particle from the bottom of the bowl is h

The particle is at a height h from the bottom of the bowl. With the rotation of the bowl, the particle also rotates about the vertical axis along a circular path of radius r.

In this situation, the weight mg of the particle of mass m, the centrifugal force mω²r and the normal force (R) on the particle by the surface of the bowl keep the particle in equilibrium.

So, \(R \sin \theta=m \omega^2 r\) and \(R \cos \theta=m g\)

∴ \(\tan \theta=\frac{\omega^2 r}{g}\)

If the radius of the bowl is a, then we get \(\tan \theta=\frac{r}{a-h}\)

∴ \(\frac{r}{a-h}=\frac{\omega^2 r}{g} \quad \text { or, } \quad a-h=\frac{g}{\omega^2}\)

or, \(h=a-\frac{g}{\omega^2}=0.1-\frac{9.8}{\omega^2}=0.1\left(1-\frac{98}{\omega^2}\right) \mathrm{m} .\)

Example 3. The radius of the earth is 6400 km. What will be the value of the centrifugal acceleration at the equatorial region due to the earth’s diurnal motion?
Solution:

Given

The radius of the earth is 6400 km.

If the radius of the earth is R and its angular velocity is ω, then at the equatorial region, centrifugal acceleration

= \(\omega^2 R=\left(\frac{2 \pi}{T}\right)^2 \times R=\left(\frac{2 \pi}{24 \times 60 \times 60}\right)^2 \times 6400 \times 10^3\)

= \(0.0338 \mathrm{~m} \cdot \mathrm{s}^{-2} .\)

Class 11 Physics Unit 3 Laws Of Motion Chapter 3 Circular Motion

Comparison Between Linear And Rotational Motion Numerical Examples

Understanding Pseudo Force: Class 11 Physics

In the discussion of rotational motion, some of the physical quantities are the rotational analogues of the corresponding physical quantities of linear motion. These quantities are given below:

Example 1. A round table Is rotated with an angular velocity of 10 rad · s^-1 about Its axis. Two blocks of mass m₁ = 10 kg and m₂ = 5 kg, connected to each other by a weightless inextensible string of length 0. 3 m, are placed along the diameter of the table. The coefficient of friction between the table and m₁ is 0. 5, while there is no friction between m₂ and the table. Mass m₁ is at a distance of 0.124 m from the centre of the table. The masses are at rest with respect to the table.

1. Calculate the frictional force on m₁.
2. What should be the minimum angular speed of the table so that these masses will slip from 1 the table?
3. How should these masses be kept so that the string remains taut but no frictional force; acts on m₁?

Solution:

Given

A round table Is rotated with an angular velocity of 10 rad · s^-1 about Its axis. Two blocks of mass m₁ = 10 kg and m₂ = 5 kg, connected to each other by a weightless inextensible string of length 0. 3 m, are placed along the diameter of the table. The coefficient of friction between the table and m₁ is 0. 5, while there is no friction between m₂ and the table. Mass m₁ is at a distance of 0.124 m from the centre of the table.

The top view of the table is shown.

According to the problem,

AB = 0.3 m;

OA = 0.124 m

So, OB = 0.3 – 0.124 = 0.176 m

1. Centrifugal force acting on mass m₁ along the direction OA, m₁ω²r = 10 x (10)² x 0.124 = 124 N

Again, centrifugal force acting on mass m₂ along the direction OB, m₂ω²r = 5 x (10)² x 0.176 = 88 N

So, the resultant force along the direction OA = 124 – 88 = 36 N

In spite of this resultant force, the two masses remain stationary with respect to the table.

Hence, the frictional force is equal and opposite to this force. Hence, the required frictional force = 36 N.

2. Reaction force of the table on mass m₁ = weight of the mass = m₁g = 10 x 9.8 = 98 N

As the coefficient of friction is 0.5, the limiting frictional force = 0.5 x 98 =49 N.

If the minimum angular velocity of the table is ω, the resultant of the two opposite centrifugal forces acting on the two masses

= m₁ω²r₁ – m₂ω²r₂ = ω²(10 x 0.124 – 5 x 0.176)

= ω²(1.24-0.88)= 0.36ω² N

According to the problem, the resultant of the two centrifugal forces = limiting frictional force

∴ 0.36 \(\omega^2=49 \text { or, } \omega^2=\frac{4900}{36}\)

or, \(\omega=\frac{70}{6}=11.67 \mathrm{rad} \cdot \mathrm{s}^{-1}\)

3. If there is no frictional force acting on the masses, the resultant of the two centrifugal forces should be zero. In that case, if mass m₁ is placed at a distance r from the centre, then the distance of m₂ becomes (0.3-r).

So, for any value of ω, \(m_1 \omega^2 r=m_2 \omega^2(0.3-r) \quad \text { or, } 0.3-r=\frac{m_1}{m_2} r\)

or, \(0.3-r=2 r \quad or, 3 r=0.3 \quad or, \quad r=0.1 \mathrm{~m}\)

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Example 2. What should be the maximum speed of a motor car of mass 2000 kg when it takes a circular turn of radius 100 m on a plane road? Coefficient of friction between the tyre and the road = 0.25. Given, g = 10 m· s^-2.
Solution:

Let the maximum speed of the motor car = v; mass of the car = m, radius of the circular path = r; coefficient of friction between the tyre and the road = μ.

In the case of maximum speed of the car, centripetal force = limiting frictional force mv2

or, \(\frac{m v^2}{r}=\mu m g\)

or, v \(=\sqrt{\mu r g}=\sqrt{0.25 \times 100 \times 10}=15.81 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Example 3. The roadway bridge over a stream is in the form of an arc of a circle of radius r. A car is crossing the bridge with a speed v. Prove that the limiting speed of the car with which it can cross the bridge without leaving the ground at the highest point of the bridge is v ≤ √gr.
Solution:

Given

The roadway bridge over a stream is in the form of an arc of a circle of radius r. A car is crossing the bridge with a speed v.

Let the mass of the car be m and the speed of the car be v.

∴ Weight of the car = mg and centrifugal force = \(\frac{m v^2}{r}\)

At the highest point on the bridge, the car will not jump up from the road if the weight of the car ≥ centrifugal force.

∴ mg ≥ \(\frac{m v^2}{r}\) or, gr ≥ \(v^2\)

or, \(v^2\) ≤ gr  or, ν ≤ \(\sqrt{g r}\).

Example 4. What will be the angular velocity of the diurnal motion of the earth, so that the weight of a body at the equatorial region is 0.6 of the present weight? Radius of the earth = 6400 km.
Solution:

Due to the diurnal motion of the earth, a centrifugal force acts on a body on the surface of the earth and hence the body suffers an apparent loss of weight.

Apparent weight of the body at the equatorial region = mg- mω²R

(ω = angular velocity, R = radius of the earth)

According to the problem, 0.6 mg = mg- mω²R

or, \(0.4 g=\omega^2 R\)

or, \(\omega=\sqrt{\frac{0.4 \times g}{R}}=\sqrt{\frac{0.4 \times 9.8}{6400 \times 1000}}=7.8 \times 10^{-4} \mathrm{rad} \cdot \mathrm{s}^{-1}\)

Short Answer Questions on Pseudo Forces

Example 5. A cylindrical drum made of steel and of diameter 20 cm is rotating about its own vertical axis. A small body made of steel remains stuck inside the cylinder when the drum rotates at the rate of 200 rpm. When the velocity of rotation decreases, the body falls down. What is the coefficient of friction between the body and the surface of the drum?
Solution:

Given

A cylindrical drum made of steel and of diameter 20 cm is rotating about its own vertical axis. A small body made of steel remains stuck inside the cylinder when the drum rotates at the rate of 200 rpm. When the velocity of rotation decreases, the body falls down.

Let the point where the body remains stuck be P

For the equilibrium of the body, weight of the body = limiting frictional force or, mg = F = μR [R = normal force on the body by the wall of the drum]

Again, R supplies the necessary centripetal force to the body for its revolution.

Hence, R = \(\frac{m v^2}{r}=m \omega^2 r\)

∴ \(m g=\mu m \omega^2 r\)

or, \(\mu=\frac{g}{\omega^2 r}=\frac{980 \times 9}{(20 \pi)^2 \times 10}\) (because \(\omega=\frac{200 \times 2 \pi}{60}=\frac{20 \pi}{3}\))

= 0.2234

Example 6. Four balls of mass 5 kg each are placed on the top of a horizontal turntable and fastened together with four strings of length 1 m each to form a square of side 1 m. The axis of rotation passes through the centre of the square. Find the tension in the strings when the turntable is rotated at the rate of \(\frac{30}{\pi}\) rpm.
SolutiIton:

Given

Four balls of mass 5 kg each are placed on the top of a horizontal turntable and fastened together with four strings of length 1 m each to form a square of side 1 m. The axis of rotation passes through the centre of the square.

The table is shown from the top. When the turntable keeps on rotating, the centrifugal force experienced by each ball will be mω²r.

Here, \(m=5 \mathrm{~kg}, \omega=2 \pi \times \frac{30}{\pi \times 60}\) = \(1 \mathrm{rad} \cdot \mathrm{s}^{-1}\)

As the length of each side of the square is 1m,

r = \(\frac{1}{2} \times \text { length of the diagonal or, } r=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}} \mathrm{~m}\)

According to the figure, tension in each string, T = \(m \omega^2 r \cos 45^{\circ}=5 \times(1)^2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{5}{2}=2.5 \mathrm{~N} \text {. }\)

Example 7. A hemispherical bowl of radius r is rotated with an angular velocity ω about a vertical axis passing through Its centre. An object rotates with the same angular velocity remaining attached to the inner surface of the bowl. The straight line Joining the body and the centre of the bowl makes an angle 45° with the vertical. The coefficient of friction between the body and the bowl is 0.2. Assuming the body is just about to move down along the curved surface of the bowl, prove that ω²r = 0.943 g.
Solution:

Given

A hemispherical bowl of radius r is rotated with an angular velocity ω about a vertical axis passing through Its centre. An object rotates with the same angular velocity remaining attached to the inner surface of the bowl. The straight line Joining the body and the centre of the bowl makes an angle 45° with the vertical. The coefficient of friction between the body and the bowl is 0.2. Assuming the body is just about to move down along the curved surface of the bowl,

The hemisphere is rotated about the vertical axis AO with an angular velocity ω. The forces acting on the body at point P are shown.

When the body is just about to move down, mg – ncosθ + μnsinθ

or, mg = n(cosθ + μsinθ) ….(1)

and mω²rsinθ = nsinθ – μncosθ

or, mω²r = n(1 -μcotθ)…(2)

From (1) and (2), we get \(\frac{\omega^2 r}{g}=\frac{1-\mu \cot \theta}{\cos \theta+\mu \sin \theta}\)

= \(\frac{1-0.2 \cot 45^{\circ}}{\cos 45^{\circ}+0.2 \sin 45^{\circ}}\left[because \mu=0.2 \text { and } \theta=45^{\circ}\right]\)

∴ \(\omega^2 r=0.943 \mathrm{~g}\)

Example 8. A car of mass m is moving with a velocity v over a bridge. What will be the values of the force Fat the highest point of a convex bridge and at the lowest point of a concave bridge?
Solution:

Given

A car of mass m is moving with a velocity v over a bridge.

1. A centripetal force is required to move the car over a convex bridge. The value of this centripetal force is mv²

Here, r is the radius of curvature of the convex bridge.

At the highest point of the bridge, the resultant of the weight of the car mg and F will supply the necessary centripetal force.

∴ \(m g-F=\frac{m v^2}{r}\)

or, F = \(m\left(g-\frac{v^2}{r}\right)\)

2. A centripetal force is required to move the car over a concave bridge also. At the lowest point of the concave bridge, the resultant of the weight of the car mg and F will supply the necessary centripetal force.

∴ \(F-m g=\frac{m v^2}{r} \quad \text { or, } F=m\left(g+\frac{v^2}{r}\right)\)

Example 9. The roadway bridge over a stream is in the form of an arc of a circle of radius 50 m. What is the maximum speed with which a car can cross the bridge without leaving the ground at the highest point?
Solution:

Given

The roadway bridge over a stream is in the form of an arc of a circle of radius 50 m.

Suppose the car is moving with a speed v on the bridge so that it can cross the bridge without leaving the ground at the highest point.

If the normal force at the highest point on the bridge is F, then \(m g-F=\frac{m v^2}{r}\)

For the maximum speed of the car, at the highest point on the bridge, F = 0

∴ \(\frac{m v^2}{r}=m g \quad \text { or, } \quad v^2=r g\)

or, \(v=\sqrt{r g}=\sqrt{50 \times 9.8}\)

= \(22 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Example 10. A car passes over a convex bridge. The centre of gravity of the car follows an arc of a circle of radius 30 m. Assuming that the car has a mass of 1000 kg, find the force that the car will exert at the highest point on the bridge if the velocity of the car is 15 m · s^-1. At what speed will the car lose contact with the road?
Solution:

Given

A car passes over a convex bridge. The centre of gravity of the car follows an arc of a circle of radius 30 m. Assuming that the car has a mass of 1000 kg, find the force that the car will exert at the highest point on the bridge if the velocity of the car is 15 m · s^-1.

Mass of the car, m = 1000 kg,

velocity, v = 15 m · s^-1

and radius of the circular path, r = 30 m

The weight of the car = mg and the necessary centripetal force =c \(\frac{m v^2}{r}\)

Let at the highest point on the convex bridge, the effective normal force by the road be F.

Then F= \(m g-\frac{m v^2}{r}=m\left(g-\frac{v^2}{r}\right)=10^3\left(9.8-\frac{15^2}{30}\right)\)

= \(2.3 \times 10^3 \mathrm{~N}\)

(because g = \(9.8 \mathrm{~m} \cdot \mathrm{s}^{-2}, v=15 \mathrm{~m} \cdot \mathrm{s}^{-1} \text { and } r=30 \mathrm{~m}\))

When F = 0, the car loses contact with the road. In that situation, if the speed of the car is u, then

⇒ \(m g-0=\frac{m u^2}{r} \text { or, } u^2=g r\)

∴ u = \(\sqrt{9.8 \times 30}=17.1 \mathrm{~m} \cdot \mathrm{s}^{-1} .\)

Pseudo Force Formula and Applications

Example 11. A spotlight is rotating with a uniform angular velocity of 0.1 rad · s^-1 on a horizontal plane. The light spot moves on a wall at a distance of 3m. If the path of light inclines at an angle of 45° with the wall, then calculate the velocity of the light-spot.

Solution:

Given

A spotlight is rotating with a uniform angular velocity of 0.1 rad · s^-1 on a horizontal plane. The light spot moves on a wall at a distance of 3m. If the path of light inclines at an angle of 45° with the wall,

Angular velocity of the spotlight S, ω = 0.1 rad · s^-1

Suppose at some moment, the light spot is at point L on the wall

∴ SL = r = \(\frac{3}{\sin 45^{\circ}}=3 \sqrt{2} \mathrm{~m}\)

If the linear velocity of the light-spot is v [v = ωr], then the velocity of the light-spot along the wall

= component of v along the wall

= \(v \sin 45^{\circ}=\frac{\omega r}{\sqrt{2}}=\frac{0.1 \times 3 \sqrt{2}}{\sqrt{2}}=0.3 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Example 12. A piece of stone of mass 1 kg is tied with a thread of length 10 m and is rotating along a horizontal circular path making an angle θ with the vertical. The thread can withstand a maximum tension of 12 N. What is the maximum velocity with which the stone can be rotated without snapping the thread?
Solution:

Given

A piece of stone of mass 1 kg is tied with a thread of length 10 m and is rotating along a horizontal circular path making an angle θ with the vertical. The thread can withstand a maximum tension of 12 N.

Let the length of the thread be l, mass and maximum linear velocity of the stone be m and v respectively and the maximum tension in the thread be T while rotating along a horizontal plane.

The required centripetal force for rotation, \(\frac{m v^2}{l \sin \theta}=T \sin \theta\)…(1)

and \(m g=T \cos \theta\)…(2)

From (1) and (2), we get, \(\sin ^2 \theta+\cos ^2 \theta=\frac{m v^2}{l T}+\left(\frac{m g}{T}\right)^2\)

or, \(\frac{m v^2}{l T}=\left(1-\frac{m^2 g^2}{T^2}\right) or, v=\left\{\frac{l T}{m}\left(1-\frac{m^2 g^2}{T^2}\right)\right\}^{1 / 2}\)

Given, \(l=10 \mathrm{~m}, T=12 \mathrm{~N}\) and \(m=1 \mathrm{~kg}\)

∴ v = \(\left\{\frac{10 \times 12}{1}\left(1-\frac{1^2 \times 9.8^2}{12^2}\right)\right\}^{1 / 2}\) (because \(g=9.8 \mathrm{~m} \cdot \mathrm{s}^{-2}\))

or, \(\nu=6.32 \mathrm{~m} \cdot \mathrm{s}^{-1}\)