ICSE CLASS X Boards Revison MCQs with Solutions | Physics WORK,POWER,ENERGY Chapter 2 | 2021 Semester 1

 

 

 

1 . A catapult throws a stone of mass 0.10 kg with a velocity of 30 m/s. If 25% of the P.E. of the elastic band is wasted during transmission, find the magnitude of the potential energy.

2. In a hydroelectric power station, 1000 kg of water is allowed to drop a height of 100 min 1 s. If the conversion of potential energy to electric energy is 60%, calculate the power output.

3. The heart of a deer chased by a tiger beats 200 times in a minute and does a work of 1.4 joules per beat. What is the power of heart?

4. An accelerated electron has energy of 9.6 * 10^(-19) J. Express energy in electron volts (eV).

 

 

 

 

 

 

 

ICSE CLASS X Boards Revison MCQs with Solutions | Physics Force Chapter 1 | 2021 Semester 1

 

 

 PAPERSOLUTIONS

 

 1.Calculate the force which will produce a moment of force of 1575 dyne cm, when the perpendicular distance between point of application of force and turning point is 45 cm.

 

2. A uniform metre scale is balanced at 60 cm mark, when weights of 5 gf and 40 gf are suspended at 10 cm mark and 80 cm mark respectively. Calculate the weight of the metre scale.

 

3 A uniform metre scale of weight 50 gf is balanced at the 40 cm mark, when a weight of 100 gf is suspended at the 5 cm mark. Where must a weight of 80 gf be suspended to balance the metre scale ?

 

4. Calculate the length of the arm of couple, if a force of 13 N produces a moment of couple of 14.3 Nm.

 

 

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 SIMPLE HARMONIC MOTION - NUMERICALS - MAHARASHTRA BOARD - SOLUTIONS - CHAPTER 5 - OSCILLATIONS

 SOLUTIONSPART1PDF

SOLUTIONSPART2PDF 

LECTURENOTES_YOUTUBE

 

 

CLASS 12 PHYSICS MAHARASHTRA BOARD CHAPTER 5 SOLUTIONS



Q1. (i) A particle performs linear S.H.M. starting from the mean position. Its amplitude is A and time period is T. At the instance when its speed is half the maximum speed, its displacement x is

(ii) A body of mass 1 kg is performing linear S.H.M. Its displacement x (cm) at t (second) is given by x = 6 sin (100t + m/4). Maximum kinetic energy of the body is

Q8. At what distance from the mean position is the speed of a particle performing S.H.M. half its maximum speed. Given path length of S.H.M. = 10 cm. [Ans: 4.33 cm]

Q9. In SI units, the differential equation of an S.H.M. d^2x—dt^2= —36x . Find its frequency and period. [Ans: 0.955 Hz, 1.05 s]

Q10. A needle of a sewing machine moves along a path of amplitude 4 cm with frequency 5 Hz. Find its acceleration (1/30) s after it has crossed the mean 30 position. [Ans: 34.2 m/s2]

Q11. Potential energy of a particle performing linear S.H.M is 0.1 pi^2 x^2 joule. If mass of the particle is 20 g, find the frequency of S.H.M. [Ans: 1.581 Hz]

Q12. The total energy of a body of mass 2 kg performing S.H.M. is 40 J. Find its speed while crossing the centre of the path. [Ans: 6.324 cm/s] 

13. A simple pendulum performs S.H.M of period 4 seconds. How much time after crossing the mean position, will the displacement of the bob be one third of its amplitude. [Ans: 0.2163 s] 

 

14. A simple pendulum of length 100 cm performs S.H.M. Find the restoring force acting on its bob of mass 50 g when the displacement from the mean position is 3 cm. [Ans: 1.48 x 10-2N] 

 

 

15. Find the change in length of a second's pendulum, if the acceleration due to gravity at the place changes from 9.75 m/s2 to 9.8 m/s2. [Ans: Decreases by 0.0051 m] 

 

 

16. At what distance from the mean position is the kinetic energy of a particle performing S.H.M. of amplitude 8 cm, three times its potential energy? [Ans: 4 cm] 

 

 

17. A particle performing linear S.H.M. of period 27π seconds about the mean position 0 is observed to have a speed of b SQRT(3) m/ s, when at a distance b (metre) from 0. If the particle is moving away from 0 at that instant, find the time required by the particle, to travel a further distance b. [Ans: π/3 s] 

18. The period of oscillation of a body of mass m1 suspended from a light spring is T. Whcn a body of mass m2 is tiEd to the first body and the system is made to oscillate, the period is 2T. Compare the masses m1 and m, [Ans: 1/3]

Q19. The displacement of an oscillating particle is given by x = asincot +bcoscot where a, b and w are constants. Prove that the particle performs a linear S.H.M. with amplitude A=sq root (a^2 +b^2)

Q20. Two parallel S.H.M.s represented by xi = 5sin (47 pi t + pi/3) cm and x2 = 3sin (4 pi t + pi/4) cm are superposed on a particle. Determine the amplitude and epoch of the resultant S.H.M. [Ans: 7.936 cm, 54° 23']

21. A 20 cm wide thin circular disc of mass 200 g is suspended to a rigid support from a thin metallic string. By holding the rim of the disc, the string is twisted through 60° and released. It now performs angular oscillations of period 1 second. Calculate the maximum restoring torque generated in the string under undamped conditions.  [Ans: 0.04133 N m] 

 

22. Find the number of oscillations performed per minute by a magnet is vibrating in the plane of a uniform field of 1.6 x 10-5 Wb/m^2. The magnet has moment of inertia 3 x 10^-6 kg/m2 and magnetic moment 3 A m^2. [Ans:38.19 osc/min.] ^

 

 

23. A wooden block of mass m is kept on a piston that can perform vertical vibrations of adjustable frequency and amplitude. During vibrations, we don't want the block to leave the contact with the piston. How much maximum frequency is possible if the amplitude of vibrations is restricted to 25 cm? In this case, how much is the energy per unit mass of the block? [Ans: nmax = 1/s, E/m = 1.25 J/kg] 

 

 

 

 

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Congruent Triangles | Class 7, Class 8 Class 9 | CBSE ICSE IGCSE

 Congruent Triangles | Class 7, Class 8 | CBSE ICSE IGCSE

 

 

 

LINKNOTES

CLASS9_NCERT_CHAPTER7 

 

CLASS8_NCERT_CHAPTER7 

Below topics are covered in this lecture: 00:00 What is Congruency? 07:50 Rules of Congruency 12:40 Altitude, Perpendicular 15:30 Median 16:42 Angle Bisector 18:10 Side bisectors 18:45 Isoceles triangle 19:14 Scalene triangle 19:48 Right Triangle 21:00 SSS 29:10 SAS 35:50 ASA 38:45 Ex7.2solutions 51:07 RHS 

 

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Ex 7.2 Class 9 Maths 

Question 1. In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at 0. Join A to 0. Show that (i) OB = OC (ii) AO bisects ∠A 

Ex 7.2 Class 9 Maths Question 2. In ∆ABC, AD is the perpendicular bisector of BC (see figure). Show that ∆ ABC is an isosceles triangle in which AB = AC. 

Ex 7.2 Class 9 Maths Question 3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see figure). Show that these altitudes are equal. 

Ex 7.2 Class 9 Maths Question 5. ABC and DBC are isosceles triangles on the same base BC (see figure). Show that ∠ ABD = ∠ACD.  

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Class 9 NCERT Solutions- Chapter 7 Triangles – Exercise 7.1
Question 1. In quadrilateral ACBD AC = AD and AB bisects ∠ A (see Fig. 7.16). Show that ∆ ABC ≅ ∆ ABD What can you say about BC and BD?
Question 3. AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.
Question 2. ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see Fig. 7.17). Prove that
(i) ∆ ABD ≅ ∆ BAC
(ii) BD = AC
(iii) ∠ ABD = ∠ BAC.

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 Simple Harmonic Motion | Derivations of x,v,a

 

Below topics have been covered in this lecture: Conditions for Linear SHM, Simple Harmonic Motion Key Terms, Mean Position, Amplitude in SHM, Time Period and Frequency of SHM, Phase in SHM, Phase Difference, Simple Harmonic Motion Equation and its Solution, Solutions of Differential Equations of SHM, Quantitative Analysis of SHM, Velocity of a particle executing Simple Harmonic Motion, Kinetic Energy of a Particle in SHM, SHM as a Projection of Circular Motion

 

 x=displacement from mean positions, v=veleocity at position x, a=acceleration at position x, ϕ- initial phase angle, w=angular frequency, k=spring constant, A=amplitude 

 

TIMESTAMP 00:00 overview (x,v,a, KE,PE, TE) 03:58 k-Spring Constant 05:06 equations overview 05:53 w,T,f discussion 11:52 how to study SHM 13:15 definition of SHM 18:47 derivation starts 23:45 Mean and Extreme values 27:45 v^2=w^2(a^2-x^2) 33:14 x=Asin(ωt+ϕ) calculations 35:00 ϕ discussion 47:14 v=Awcos(wt+ϕ) calculations 48:30 a=-Aw^2sin(ωt+ϕ) calculations 49:10 T calculations 

 

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#isequaltoklasses #shm #simpleharmonicmotion