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## Numericals from Newton’s Laws of motion based on F=ma (For calss XI students of Kendriya Vidyalaya Pattom)

Download the questions and solve them and submit on or before 9 Oct 2014

1. A force acts for 10 s on a body of mass 10 kg after which the force ceases and the body describes 50 m in the next 5 s. Find the magnitude of the force. [Ans: 10 N]
2. A  truck starts from rest and rolls down a hill with constant acceleration. It travels a distance od 400 m in 20s. Calculate the acceleration and the force acting on it if its mass is 7 metric tonnes. [Ans: 2 m/s2, 14000N]
3. A motor car running aat the rate of 7 m/s can be stopped by applying brakes in 10 m. Show that total resistance to the motion, when the brakes are on is one fourth of the weight of the car.
4. In an Xray machine, an electron is subjected top a force of 10-23 N. In how much time the electron will cover a distance of 0.1 m,? Take the mass of the electron = 10-30 kg

## Numericals from Newton’s Laws of motion based on F=ma (For calss XI students of Kendriya Vidyalaya Pattom)

Download the questions and solve them and submit on or before 9 Oct 2014

1. A force acts for 10 s on a body of mass 10 kg after which the force ceases and the body describes 50 m in the next 5 s. Find the magnitude of the force. [Ans: 10 N]
2. A  truck starts from rest and rolls down a hill with constant acceleration. It travels a distance od 400 m in 20s. Calculate the acceleration and the force acting on it if its mass is 7 metric tonnes. [Ans: 2 m/s2, 14000N]
3. A motor car running aat the rate of 7 m/s can be stopped by applying brakes in 10 m. Show that total resistance to the motion, when the brakes are on is one fourth of the weight of the car.
4. In an Xray machine, an electron is subjected top a force of 10-23 N. In how much time the electron will cover a distance of 0.1 m,? Take the mass of the electron = 10-30 kg

Mr. MS Kumaraswamy, a trained graduate teacher in Kendriya Vidyalaya Donimalai has prepared an excellent set of question banks and worksheets in Physics and maths for classes VI to X.\You can doenload the Physics question bank at the following Link. ## Questions from Alternating current

1.      Derive an expression for the average power in an ac circuit.
2.      Distinguish between resistance, reactance and impedance.
3.      Draw the phasor diagram showing voltage and current in LCR series circuit and derive an expression for the impedance 4.      Deduce the phase relationship between current and voltage in an ac circuit containing inductor only
5.      A series LCR circuit with L = 0.12 H, C = 4.8 x 10 7 f and R = 23 ohms is connected to a variable frequency supply.  At What frequency is the current maximum?
9.      A sinusoidal voltage V=200 sin 314t is applied to a resistor of 10 ohms resistance. Calculate (i) rms value of the voltage, (ii) rms value of the current and (iii) power dissipated as heat in watts.

(Posted by Sayantika Nath) ## Kendriya Vidyalaya Sangathan invites application for various posts

Kendriya Vidyalaya Sangathan invites application for various teaching and miscellaneous posts. The details can be had from http://kvsangathan.nic.in/EmploymentDocuments/emp-19-07-13e.pdf

KENDRIYA VIDYALAYA SANGATHAN – a premier organization in India administering1093 schools as on 01.05.2013 known as ”Kendriya Vidyalayas” with 11,21,012 students as on 31.03.2013 and 56,445 employees on rolls as on 01.10.2012. (more…)

## Rotational Equilibrium – A Numerical Problem

A father and his son are supposed to carry a load of 500 N. They decided to use a uniform pole that is 4.0 m long and weighs 60 N. The father and his son support the pole at the ends. Where should the load be placed so that the father supports three times as much as his son?

Posted by Daniel

Let’s represent the situation be the folowing diagram Let the force exerted by son be f

Then the force exerted by father = 3f

Therefore the total force exerted =Total weight = 560N = 4f

Which gives f = 140 N, the force exerted by son

3f=420 N, the force exerted by father

Applying the principle of moments about the point where father holds the rod,

420 x 0 – 500 x x – 60 x 2 + 140 x 4 = 0

==> x = 440/500 = 0.88 m from the end where father holds the load

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