HC Verma Concepts Of Physics Exercise Solutions Of Chapter 44 (X-Rays)

HC Verma Concepts of Physics Solutions - Part 1, Chapter 44 - X-Rays:

EXERCISE

Planck constant, h = 4.14 * 10^–15 eV-s or, 6.63 * 10^–34 J-s; speed of light, c = 3 * 10⁸ m/s.

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1. Find the energy, the frequency and the momentum of an X-ray photon of wavelength 0.10 nm.

Sol:

Given: wavelength, λ = 0.10 nm = 1 * 10^–10 m.

We know,

→ λ = c/ν

Or, ν = c/λ

Or, ν = (3 * 10⁸)/ (1 * 10^–10)

Or, ν = 3 * 10¹⁸ s^–1 or Hz.

So, the frequency of wavelength is 3 * 10¹⁸ s^–1 or Hz.

→ Energy, E = hν

Or, E = 4.14 * 10^–15 * 3 * 10¹⁸

Or, E = 12.4 * 10³ eV = 12.4 keV.

→ Momentum, p = E/c

Or, p = (12.4 * 10³ * 1.6 * 10^–19)/ (3 * 10⁸)

Or, p = 6.62 * 10^–24 kg m/s.

2. Iron emits K_α X-ray of energy 6.4 keV and calcium emits K_α X-ray of energy 3.69 keV. Calculate the times taken by an iron K_α photon and a calcium K_α photon to cross through a distance of 3 km.

Sol:

Given: distance, S = 3 km = 30000 m.

Speed of both the photon are same and equal to speed of light, c = 3 * 10⁸ m/s.

So, the times taken by both = S/c = 3000/ (3 * 10⁸) = 10 µs.

3. Find the cut-off wavelength for the continuous X-rays coming from an X-ray tube operating at 30 kV.

Sol:

Given: potential, V = 30 kV = 30000 V.

We know,

→ λ_min = hc/eV

Or, λ_min = (4.14 * 10^–15 eV-s* 3 * 10⁸ m/s)/ (30000 eV)

Or, λ_min = 41.4 * 10^–12 m = 41.4 pm.

4. What potential difference should be applied across an X-ray tube to get X-ray of wavelength not less than 0.10 nm? What is the maximum energy of a photon of this X-ray in joule?

Sol:

Given: λ_min = 0.10 nm = 0.1 * 10^–9 m.

We know,

→ λ_min = hc/eV

Or, V = hc/ (eλ_min)

Or, V = (4.14 * 10^–15 eV-s * 3 * 10⁸ m/s)/ (e * 0.1 * 10^–9)

Or, V = 12.4 * 10³ V = 12.4 kV.

→ Maximum energy, E = eV

Or, E = 1.6 * 10^–19 * 12.4 * 10³

Or, E = 1.98 * 10^–15 J.

5. The X-ray coming from a Coolidge tube has a cut-off wavelength of 80 pm. Find the kinetic energy of the electrons hitting the target.

Sol:

Given: λ = 80 pm = 0.8 * 10^–10 m.

We know,

→ λ = hc/E

Or, E = hc/λ

Or, E = (4.14 * 10^–15 * 3 * 10⁸)/ (0.8 * 10^–10)

Or, E = 15.5 * 10³ eV = 15.5 keV.

So, the kinetic energy of the electrons is 15.5 keV.

6. If the operating potential in an X-ray tube is increased by 1%, by what percentage does the cutoff wavelength decrease?

Sol:

We know,

Wave length, λ = hc/eV

Now, potential, V is increased by 1%.

New wave length, λ’ = hc/1.01eV = λ/1.01

→ Δλ = λ – λ’ = (0.01/1.01) λ

So, % change of wave length = (Δλ/λ)*100

Or, % change of wave length = 1/1.01

Or, % change of wave length = 0.99 = 1 %.

7. The distance between the cathode (filament) and the target in an X-ray tube is 1.5 m. If the cutoff wavelength is 30 pm, find the electric field between the cathode and the target.

Sol:

Given: distance, d = 1.5 m; wavelength, λ = 30 pm = 0.3 * 10^–10 m. 

We know, E = hc/λ

Or, E = (4.14 * 10^–15 * 3 * 10⁸)/ (0.3 * 10^–10)

Or, E = 41.4 * 10³ eV

Or, potential, V = E/e = 41.4 * 10³ V

Now, Electric field = V/d

Or, Electric field = 41.4 * 10³/1.5

Or, Electric field = 27.6 * 10³ V/m

Or, Electric field = 27.6 kV/m.

8. The short-wavelength limit shifts by 26 pm when the operating voltage in an X-ray tube is increased to 1.5 times the original value. What was the original value of the operating voltage?

Sol:
Given: λ’ = λ – 26 * 10^–12; V’ = 1.5 V.

Original wavelength, λ = hc/eV

New wavelength, λ’ = hc/eV’

→ λV = λ’V’

Or, λV = (λ – 26 * 10^–12) * 1.5 V

Or, λ = (1.5 * 26 * 10^–12)/0.5

Or, λ = 78 * 10^–12 m.

So, original voltage, V = hc/eλ

Or, V = (4.14 * 10^–15 eV-s * 3 * 10⁸ m/s)/ (e * 78 * 10^–12)

Or, V = 15.93 * 10³ V = 15.93 kV.

9. The electron beam in a colour TV is accelerated through 32 kV and then strikes the screen. What is the wavelength of the most energetic X-ray photon?

Sol:

Given: voltage, V = 32 kV = 32000 V.

We know, λ = hc/eV

Or, λ = (4.14 * 10^–15 eV-s* 3 * 10⁸ m/s)/ (320000 eV)

Or, λ = 38.8 * 10^–12 m = 38.8 pm.

10. When 40 kV is applied across an X-ray tube, X-ray is obtained with a maximum frequency of 9.7 * 10¹⁸ Hz. Calculate the value of Planck constant from these data.

Sol:

Given: voltage, V = 40 kV = 40 * 10³ V; frequency, ν = 9.7 * 10¹⁸ Hz.

We know, E = hν

Or, eV = hν

Or, h = eV/ν

Or, h = 40 * 10³/9.7 * 10¹⁸ eV-s

Or, h = 4.12 * 10^–15 eV-s.

11. An X-ray tube operates at 40 kV. Suppose the electron converts 70% of its energy into a photon at each collision. Find the lowest three wavelengths emitted from the tube. Neglect the energy imparted to the atom with which the electron collides.

Sol:

Given: voltage, V = 40 kV = 40 * 10³ V.

Energy, E = eV = 40 * 10³ eV.

→ Energy utilized = (70/100) * 40 * 10³ eV.

Or, Energy utilized = 28 * 10³ eV.

Now, λ = hc/E

Or, λ = (4.14 * 10^–15 * 3 * 10⁸)/ (28 * 10³)

Or, λ = 44.35 * 10^–12 = 44.35 pm.

For 2^nd wavelength:

→ Energy, E = 70% of left energy

Or, E = (70/100) * (40 – 28) * 10³ eV.

Or, E = 84 * 10² eV.

Now, λ = hc/E

Or, λ = (4.14 * 10^–15 * 3 * 10⁸)/ (84 * 10²)

Or, λ = 148 * 10^–12 = 148 pm.

For 3^rd wavelength:

→ Energy, E = 70% of left energy

Or, E = (70/100) * (12 – 8.4) * 10³ eV.

Or, E = 25.2 * 10² eV.

Now, λ = hc/E

Or, λ = (4.14 * 10^–15 * 3 * 10⁸)/ (25.2 * 10²)

Or, λ = 493 * 10^–12 = 493 pm.

Discussion - If you have any Query or Feedback comment below.

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JEE Previous Year Questions With Solutions Of Newton's Laws Of Motion Part - 3 | IIT JEE | AIEEE | NEET | JEE Main & Advance | AIIMS | WBJEE | KCET | PMT | OTHER JE EXAMS

It is very important to have the idea of any examination to get success in that. The best way to have that idea is to have a look at previous year Questions of that examination. That's why we have collected some previous year Questions with Solutions of JEE Main, JEE Advance, NEET, IIT JEE, AIEEE, AIIMS, WBJEE, EAMCET, Karnataka CET, CPMT, Kerala CET, MP PMT and Other State JE Exams so that you can get clear pattern and level of question that will be asked in forthcoming JEE Main, JEE Advance, IIT JEE, AIEEE, AIIMS, WBJEE, EAMCET, Karnataka CET, CPMT, Kerala CET, MP PMT and Other State JE Exams.

JEE Previous Year's Questions With Solutions

1. A particle of mass 0.3 kg is subjected to a force F = – kx with k = 15 N/m. What will be its initial acceleration if it is released from a point 20 cm away from the origin? (AIEEE 2005)

(A) 3 m/s²

(B) 5 m/s²

(C) 10 m/s²

(D) 15 m/s².

Sol: (C)

Given: mass, m = 0.3 kg; k = 15 N/m; x = 20 cm = 0.2 m.

We have, force F = – kx

Or, F = 15 * 0.2 = 3 N.

And, acceleration = F/m = 3/0.3 = 10 m/s².

2. A block is kept on a frictionless inclined surface with angle of inclination α. The incline is given an acceleration ‘a’ to keep the block stationary. Then ‘a’ is equal to: (AIEEE 2005)

(A) g tan α

(B) g/tan α

(C) g cosec α

(D) g.

Sol: (A) 

The equation of motion of the block along the inclination (a = 0):

→ ma cos α – mg sin α = ma

Or, ma cos α – mg sin α = 0

Or, a = g tan α.

3. A player caught a cricket ball of mass 150 g moving at a rate of 20 m/s. If the catching process is completed in 0.1 s, the force of the blow exerted by the ball on the hand of the player is equal to: (AIEEE 2006)

(A) 3 N

(B) 150 N

(C) 30 N

(D) 300 N.

Sol: (C)

Given: mass, m = 150 g = 0.15 kg; initial velocity, u = 20 m/s; final velocity, v = 0; time taken, t = 0.1 s.

Acceleration, a = (v – u)/t = – 200 m/s²

So, force exerted by the ball on the hand of the player is = ma = 0.15 * 200 = 30 N.

4. A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m which applying the force and the ball goes upto 2 m height further, find the magnitude of the force. Consider g = 10 m/s²: (AIEEE 2006)

(A) 4 N

(B) 16 N

(C) 22 N

(D) 20 N.

Sol: (D)

Given: S = 0.2 m; mass, m = 0.2 kg; height, h = 2 m.

→Work done by hand = Potential energy of the ball

Or, FS = mgh

Or, F = mgh/S = (0.2 * 10 * 2)/0.2 = 20 N.

5. A block of mass m is connected to another of mass M by a spring (massless) of spring constant k. The blocks are at rest and the spring is unstretched. Then a constant force F starts acting on the block of mass M to pull it. Find the force of the block of mass m: (AIEEE 2007)

(A) mF/M

(B) MF/ (M + m)

(C) Fm/ (M + m)

(D) F (M + m)/m.

Sol: (C)

Taking M and m are together as a system. External force acting on the system is F.

Therefore, acceleration a = F/ (M + m).

So, force on the block of mass m = ma = Fm/ (M + m).

6. A 1 N pendulum bob is held at an angle θ from the vertical by a 2 N horizontal force F as shown in the figure. The tension in the string supporting the pendulum bob (in Newton) is: (EAMCET 2011)

(A) 1

(B) √5

(C) 2/cos θ

(D) Cos θ.

Sol: (B)

Given: F = 2 N; weight, mg = 1 N.

Let the tension in the string be T.

Force balance along vertical direction:

→ T cos θ = mg

Or, T cos θ = 1 ------- (1)

Force balance along horizontal direction:

→ T sin θ = F

Or, T sin θ = 2 ------- (2)

Solving equation (1) and (2)

We get, tan θ = 2

And sin θ = 2/√5, cos θ = 1/√5.

Therefore, T = √5.

7. The maximum tension a rope can withstand is 60 kg wt. The ratio of maximum acceleration with which two boys of masses 20 kg and 30 kg can climb up the rope at the same is: (EAMCET 2011)

(A) 3 : 2

(B) 4 : 3

(C) 2 : 1

(D) 1 : 2.

Sol: (C)

Given: maximum tension of rope, T = 60 kg-wt = 60g N.

The equation of motion of boy of mass 20 kg:

→ Ma₁ = T – Mg

Or, 20a₁ = 60g – 20g

Or, a₁ = 2g.

The equation of motion of boy of mass 30 kg:

→ Ma₂ = T – Mg

Or, 30a₂ = 60g – 30g

Or, a₂ = g.

So, the ratio of maximum acceleration is a₁ : a₂ = 2 : 1.

8. Choose the correct statement

(1) The position of centre of mass of a system is dependent on the choice of co-ordinate system.

(2) Newton’s second law of motion is applicable to the centre of mass of the system.

(3) When no external force acts on a body, the acceleration of centre of mass is zero.

(4) Internal forces can change the state of centre of mass. (EAMCET 2012)

(A) Both (1) and (2) are correct

(B) Both (2) and (3) are wrong

(C) Both (1) and (3) are wrong

(D) Both (1) and (4) are wrong.

Sol: (D)

Statement (1) is wrong.

Statement (2) is correct.

Statement (3) is correct.

Statement (4) is wrong.

Therefore, option (D) is correct.

10. Two wooden blocks of masses M and m are placed on a smooth horizontal surface as shown in figure. If a force P is applied to the system as shown in figure such that the mass m remains stationary with respect to block of mass M, then the magnitude of the force P is: (EAMCET 2013)

(A) g tan β

(B) mg cos β

(C) (M + m) g cosec β

(D) (M + m) g tan β.

Sol: (C)

Acceleration of the system, a = P/ (M + m).

The equation of motion of the block along the inclination (a’ = 0):

→ ma cos β – mg sin β = ma’

Or, (Pm cos β)/ (M + m) – mg sin β = 0

Or, P = (M + m) g tan α.

11. A mass M kg is suspended by a weightless string. The horizontal force required to hold the mass at 60⁰ with the vertical is: (EAMCET 2014)

(A) Mg

(B) Mg √3

(C) Mg (√3 + 1)

(D) Mg/ √3.

Sol: (B)

Let the tension in the string be T and horizontal force be F.

Force balance along vertical direction:

→ T cos 60⁰ = Mg

Or, T/2 = Mg ------- (1)

Force balance along horizontal direction:

→ T sin 60⁰ = F

Or, T √3/2 = F ------- (2)

Solving equation (1) and (2)

We get, F = Mg √3.

12. A force (2i + j – k) N acts on a body which is initially at rest. At the end of 20 s the velocity of the body is (4i + 2j – 2k) m/s, then the mass of the body is: (EAMCET 2015)

(A) 4.5 kg

(B) 5 kg

(C) 8 kg

(D) 10 kg.

Sol: (D)

Magnitude of force, F = √ [2² + 1² + (– 1)²] = √6 N.

Magnitude of final velocity, v = √ [4² + 2² + (– 2)²] = √24 m/s.

Initial velocity, u = 0; time taken, t = 20 s.

Acceleration, a = (v – u)/t = √24/20.

We know, force = mass * acceleration

Or, √6 = mass * √24/20

Or, mass = 10 kg.

13. A cricket ball of mass 0.25 kg with speed 10 m/s collides with a bat and returns with same speed within 0.01 s. The force acted on bat is: (WBJEE 2011)

(A) 500 N

(B) 50 N

(C) 250 N

(D) 25 N.

Sol: (A)

Given: mass, m = 0.25 kg; time, t = 0.01 s; change in velocity, ΔV = 10 – (–10) = 20 m/s.

Acceleration, a = ΔV/t = 20/0.01 = 2000 m/s².

Force, F = ma = 0.25 * 2000 = 500 N.

14. Three blocks of mass 4 kg, 2 kg, 1 kg respectively are in contact on a frictionless table as shown in the figure. If a force of 14 N is applied on the 4 kg block, the contact force between the 4 kg and the 2 kg block will be: (WBJEE 2012)

(A) 2 N

(B) 6 N

(C) 8 N

(D) 14 N.

Sol: (A)

Taking three blocks as a system (mass, M = 4 + 2 + 1 = 7 kg); force, F = 14 N:

Acceleration, a = F/M = 14/7 = 2 m/s².

From the F.B.D. of 4 kg block:

→ 14 – F_4,2 = ma

Or, F_4,2 = 14 – 4 * 2

Or, F_4,2 = 6 N.

15. A mass of 1 kg is suspended by means of a thread. The system is (i) lifted up with an acceleration of 4.9 m/s², (ii) lowered with an acceleration of 4.9 m/s². The ratio of tension in the first and second case is: (WBJEE 2016)

(A) 3:1

(B) 1:2

(C) 1:3

(D) 2:1

Sol: (A)

(i) The equation of motion of the block when lifted up with an acceleration of 4.9 m/s²:

→ Ma = T₁ – Mg

Or, 1 * 4.9 = T₁ – 1 * 9.8

Or, T₁ = 14.7 N.

(ii) The equation of motion of the block when lowered with an acceleration of 4.9 m/s²:

→ Ma = Mg – T₂

Or, 1 * 4.9 = 1 * 9.8 – T₂

Or, T₂ = 4.9 N.

So, the ratio of tension in the first and second case is 3:1.

Discussion - If you have any Query or Feedback comment below.

Tagged With: Previous year Chapter-wise JEE Questions Solutions, Chapter-wise previous year solutions of IIT JEE, AIEEE, Other JEE Exams, HC Verma Solutions, Concepts ofPhysics Volume 1, Concepts of Physics Volume 2, HCVERMA SOLUTION (CHAPTERWISE), JEE Main and advance questions and solutions.

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JEE Previous Year Questions With Solutions Of Newton's Laws Of Motion Part - 2 | IIT JEE | AIEEE | NEET | JEE Main & Advance | AIIMS | WBJEE | KCET | PMT | OTHER JE EXAMS

It is very important to have the idea of any examination to get success in that. The best way to have that idea is to have a look at previous year Questions of that examination. That's why we have collected some previous year Questions with Solutions of JEE Main, JEE Advance, NEET, IIT JEE, AIEEE, AIIMS, WBJEE, EAMCET, Karnataka CET, CPMT, Kerala CET, MP PMT and Other State JE Exams so that you can get clear pattern and level of question that will be asked in forthcoming JEE Main, JEE Advance, NEET, IIT JEE, AIEEE, AIIMS, WBJEE, EAMCET, Karnataka CET, CPMT, Kerala CET, MP PMT and Other State JE Exams.

JEE Previous Year's Questions With Solutions

1. A block of mass M is pulled along a horizontal frictionless surface by a rope of mass. If a force P is applied at the free end of the rope, the force exerted by the rope on the block is: (AIEEE 2003)

(A) Pm/ (M – m)

(B) P

(C) PM/ (M + m)

(D) Pm/ (M + m)

Sol: (C)

Taking (Block + Rope) as a system:

Acceleration of the system, a = P/ (M + m)

F_B,R = the force exerted by the rope on the block.

From the F.B.D. of the block:

→ Ma = F_B,R

Or, F_B,R = M * [P/ (M + m)]

Or, F_B,R = PM/ (M + m).

2. A rocket with a lift- off mass 3.5 * 10⁴ kg is blasted upwards with an initial acceleration of 10 m/s². Then the initial thrust of the blast is: (AIEEE 2003)

(A) 7.0 * 10⁵ N

(B) 3.5 * 10⁵ N

(C) 1.75 * 10⁵ N

(D) 14.0 * 10⁴ N.

Sol: (B)

Given: mass, m = 3.5 * 10⁴ kg; acceleration, a = 10 m/s².

→ Initial thrust, F = ma

Or, F = 3.5 * 10⁴ * 10 = 3.5 * 10⁵ N.

3. A light spring balance hangs from the hook of the other light spring balance and a block of mass M kg hangs from the former one. Then the true statements about the scale reads is: (AIEEE 2003)

(A) Both the scales read M/2 kg

(B) Both the scales read M kg each

(C) The scale of the lower one reads M kg and of the upper one zero

(D) The reading of the two scales can be anything but the sum of the reading will be M kg.

Sol: (B)

Both the scales read M kg each.

4. A machine gun fires a bullet of mass 40 g with a velocity 1200 m/s. The man holding it can exert a maximum force of 144 N on the gun. How many bullets can he fire per second at the most? (AIEEE 2004)

(A) Four

(B) Two

(C) Three

(D) One.

Sol: (C)

Given: mass of bullet, m = 40 g = 0.04 kg; final velocity, v = 1200 m/; initial velocity, u = 0.

Let he can fire n bullets per second.

We have, Force = change in momentum per second

Or, 144 = n * 0.04 * (1200 – 0)

Or, n = 3.

5. Two masses m₁ = 5 kg and m₂ = 4.8 kg tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when lift free to move? (AIEEE 2004, Odisha JEE 2002, UPSEAT 2002)

(A) 5 m/s²

(B) 0.2 m/s²

(C) 4.8 m/s²

(D) 9.8 m/s².

Sol: (C) 

The equation of motion of m₁ = 5 kg:

→ m₁g – T= m₁a

Or, 5g – T = 5a ------- (1)

The equation of motion of m₁ = 4.8 kg:

→ T – m₂g = m₂a

Or, T – 4.8g = 4.8a ------- (2)

Solving equation (1) and (2)

We get, a = 0.2 m/s².

6. A block A of mass 7 kg is placed on a frictionless table. A thread tied to it passes over a frictionless pulley and carries a body B of mass 3 kg at the other end. The acceleration of the system is (take g = 10 m/s²): (Kerala CET 2001)

(A) Zero

(B) 3 m/s²

(C) 30 m/s²

(D) 10 m/s²

(E) 100 m/s².

Sol: (B) 

Let T be the tension in the string and a be the acceleration.

The equation of motion of A:

→ T = Ma

Or, T = 7a ------- (1)

The equation of motion of B:

→ Mg – T = Ma

Or, 3g – T = 3a ------- (2)

Solving equation (1) and (2)

We get, a = 3 m/s².

7. A cricketer catches a ball of mass 150 g in 0.1 s moving with speed 20 m/s, then he experiences force of: (CBSE 2001)

(A) 0.3 N

(B) 3 N

(C) 30 N

(D) 300 N.

Sol: (C)

Given: mass of ball, m = 150 g = 0.15 kg; change of velocity, ΔV = 200 – 0 = 200 m/s; time, t = 0.1 s.

We know, Force, F = ma

Or, F = 0.15 * 200/0.1 = 30 N

8. A body of mass 1 kg is falling with an acceleration of 10 m/s². Its apparent weight will be (g = 10 m/s²): (MP PMT 2002)

(A) Zero

(B) 0.5 kg wt

(C) 2.0 kg wt

(D) 1.0 kg wt.

Sol: (A)

Given: mass, m = 1 kg; acceleration, a = 10 m/s²; g = 10 m/s².

Apparent weight, W’ = m (g – a)

Or, W’ = 1 * (10 – 10)

Or, W’ = 0.

9. A body whose mass is 50 kg stands on a spring balance inside a lift. The lift starts to ascend with an acceleration of 2 m/s². The reading of the machine or balance (g = 10 m/s²): (Kerala CET 2002)

(A) 60 kg

(B) 49 kg

(C) 50 kg

(D) 45 kg

(E) Zero.

Sol: (A)

Given: mass, m = 50 kg; acceleration, a = 2 m/s²; g = 10 m/s².

Apparent weight, W’ = m (g + a)

Or, W’ = 50 * (10 + 2)

Or, W’ = 600 N.

Now, W’ = m’g

Or, m’ = 60 kg.

10. A ball of mass 0.5 kg moving with a velocity of 2 m/s strikes a wall normally and bounces back with the same speed. If the time of contact between the ball and wall is 10^{– 3} s, the average force exerted by the wall on the ball is: (Kerala CET 2002, Haryana CEET 2002)

(A) 5000 N

(B) 1000 N

(C) 1125 N

(D) 2000 N

(E) 500 N.

Sol: (D)

Given: mass, m = 0.5 kg; time, t = 10^{– 3} s; change in velocity, ΔV = 2 – (–2) = 4 m/s.

Acceleration, a = ΔV/t = 4/10^{– 3} = 4 * 10³ m/s².

Force, F = ma = 0.5 * 4 * 10³ = 2000 N.

11. You are on a frictionless horizontal plane. How can you get off if no horizontal force is exerted by pushing against the surface? (J&K CET 2002)

(A) By running on the plane

(B) By spitting or sneezing

(C) By jumping

(D) By rolling your body on the surface.

Sol: (B)

By spitting or sneezing.

12. A 1 kg particle strikes a wall with velocity 1 m/s at an angle 30⁰ and reflects at the same angle. If it remains in contact with wall for 0.1 s, then the force is: (CBSE 2000, CPMT 2001)

(A) 10√3 N

(B) 30√3 N

(C) 40√3 N

(D) Zero.

Sol: (A) 

Horizontal motion:

Change in velocity, ΔV = [(V cos 30⁰) – (– V cos 30⁰)] = 2 * 1 * √3/2 = √3 m/s.

Acceleration, a = (√3)/ 0.1 = 10√3 m/s².

Therefore, Force F = ma = 1 * 10√3 = 10√3 N.

13. A lift of mass 1000 kg which is moving with acceleration of 1 m/s² in upward direction, then the tension developed in string which is connected to lift is: (CBSE 2002)

(A) 10800 N

(B) 11000 N

(C) 10000 N

(D) 9800 N.

Sol: (B)

Given: mass, m = 1000 kg; acceleration, a = 1 m/s².

The equation of motion of lift:

→ ma = T – mg

Or, 1000 * 1 = T – 1000 * 10

Or, T = 11000 N.

14. A ball of mass 150 g moving with acceleration of 20 m/s² is hit by a force which acts on it for 0.1 s. The impulsive force is: (Punjab PMT 2003)

(A) 0.1 Ns

(B) 1.2 Ns

(C) 0.5 Ns

(D) 0.3 Ns.

Sol: (D)

Given: mass, m = 150 g = 0.15 kg; acceleration, a = 20 m/s²; time, t = 0.1 s.

We know, Force = ma = 0.15 * 20 = 3 N.

Therefore, Impulse = Ft = 3 * 0.1 = 0.3 Ns.

15. Two masses M and M/2 are joined together by means of light inextensible string passed over a frictionless pulley as shown in figure. When the bigger mass is released, the small one will ascend with an acceleration of: (Kerala CET 2005)

(A) g/3

(B) g/2

(C) 3g/2

(D) g.

Sol: (A) 

The equation of motion of M kg block:

→ Ma = Mg – T ------- (1)

The equation of motion of m₁ = 4.8 kg:

→ Ma/2 = T – Mg/2 ------- (2)

Solving equation (1) and (2)

We get, a = g/3.

Discussion - If you have any Query or Feedback comment below.

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