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स्वागत गीत : अथ स्वागतम् शुभ स्वागतम् (Ath Swagatam Shubh Swagatam)

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स्वागत गीत : अथ स्वागतम् शुभ स्वागतम् (Ath Swagatam Shubh Swagatam)  अथ स्वागतं शुभ स्वागतम्  स्वागतम् । अथ स्वागतं शुभ स्वागतम् । आनंद मंगल मंगलम् । नित प्रियं भारत भारतम् ॥ ध्रु.॥ नित्य निरंतरता नवता मानवता समता ममता सारथि साथ मनोरथ का जो अनिवार नहीं थमता संकल्प अविजित अभिमतम् ॥ १॥ आनंद मंगल मंगलम् । नित प्रियं भारत भारतम् । अथ स्वागतं शुभ स्वागतम् ॥ कुसुमित नई कामनाएँ सुरभित नई साधनाएँ मैत्रीमात क्रीडांगन में प्रमुदित बन्धु भावनाएँ शाश्वत सुविकसित इति शुभम् ॥ २॥ आनंद मंगल मंगलम् । नित प्रियं भारत भारतम् । अथ स्वागतं शुभ स्वागतम् ॥

Ncert Solution Class 11 Physics Chapter 13 OSCILLATIONS - Param Himalaya

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13.1 Which of the following examples represent periodic motion? (a) A swimmer completing one (return) trip from one bank of a river to the other and back. (b) A freely suspended bar magnet displaced from its N-S direction and released. (c) A hydrogen molecule rotating about its centre of mass. (d) An arrow released from a bow. Solution:  (a)  It is not a periodic motion. Although the motion of the swimmer is to and from but it does not have a definite time period. (b) It is a periodic motion. Once the freely suspended magnet is displaced and is allowed to execute motion, it will oscillate about a fixed point with a definite time period. In facts , it will execute S.H.M. (c) It is a periodic motion. (d) It is not a periodic motion. 13.2 Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion? (a) the rotation of earth about its axis. (b) motion of an oscillating mercury column in a U-tube. (c) motion ...

NCERT Solution Class 11 Physics Chapter 14 Waves - Param Himalaya

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14.1 A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end? Solution:  $M \ of \ the \ string , M = 2.50 kg$ $Tension \ in \ the \ string , T = 200N$ $Length \ of \ the \ string , l = 20.0 m$ $Mass \ per \ unit \ length$ $\mu = \frac{M}{l}$ $\mu = \frac{2.50}{20}$ $\mu = 0.125 kg/m$ The velocity (v) of the transverse wave in the string is given by the relation:  $v= \sqrt{\frac{T}{\mu}}$ $v = \sqrt{\frac{200}{0.125}}$ $v = \sqrt{\frac{200 \times 1000}{125}}$ $v = \sqrt{1600}$ $v = 40 m/s$ Time taken by the disturbance to reach the other end ,  $t =\frac{l}{v}$ $t = \frac{20}{40}$ $t = 0.50 sec$ 14.2 A stone dropped from the top of a tower of height 300 m splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340 m s–1 ? (g = 9.8 m...

NCERT Solution Class 11 Physics Chapter 5 Work, Energy and Power - Param Himalaya

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NCERT Solution Class 11 Physics Chapter 5 Work, Energy and Power - Param Himalaya  5.1 The sign of work done by a force on a body is important to understand. State carefully. if the following quantities are positive or negative: (a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket. (b) work done by gravitational force in the above case, (c) work done by friction on a body sliding down an inclined plane, (d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity, (e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest. Solution :  (a) Positive: The man applies force upwards, and the bucket moves upwards. Force and displacement are in the same direction, so the work done is positive. (b) Negative: Gravitational force acts downwards, while the bucket moves upwards. Force and displacement are in opposite directions, so the work done is negative. (c) Negative: Fric...

NCERT Solution Class 11 Physics Chapter 4 Laws of Motion- Param Himalaya

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NCERT Solution Class 11 Physics Chapter 4 Laws of Motion- Param Himalaya 4.1 Give the magnitude and direction of the net force acting on (a) a drop of rain falling down with a constant speed, (b) a cork of mass 10 g floating on water, (c) a kite skillfully held stationary in the sky, (d) a car moving with a constant velocity of 30 km/h on a rough road, (e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields. Solution:  (a) Raindrop falling with constant speed:  * Magnitude: 0 N  * Direction: No direction (since the net force is zero) When an object moves with constant speed, its acceleration is zero. According to Newton's second law (F = ma), if acceleration is zero, the net force acting on the object is also zero. (b) Cork floating on water:  * Magnitude: 0 N  * Direction: No direction (since the net force is zero) The cork is in equilibrium, meaning the upward buoyant force from the water exactly balances the ...