Question

26. What is the velocity of a body of mass 100 g having K.E. of 20 J ?
(1) 2 m/s
(2) 20 m/s
(3) 40 m/s
(4) None of these
27. As the frequency of a source decreases in a given medium, the wavelength of a periodic
longitudinal wave
(1) increases, but the speed of the wave remains constant.
(2) increases, and the speed of the wave increases.
(3) decreases, but the speed of the wave remains constant.
(4) decreases, and the speed of the wave decreases.
28. What will be the value of acceleration due to gravity at a height 2R from the surface of earth?
(1) 1.1 m/s2
(2) 2.2 m/s2
(3) 3.3 m/s2
(4) 4.9 m/s2
​

Answer 1

$\huge\mathfrak{Answer :}$

__________________________________

$<b><I>QUESTION 26.$

Given,

Mass = 100 g = 0.1 kg

K.E. = 20J

Velocity = ?

Kinetic Energy = 1/2 × mass × (velocity)^2

20 = 1/2 × 0.1 × v^2

20 × 2 / 0.1 = v^2

400 = v^2

20ms^-1 = v

ANSWER : Option 2

QUESTION 27.

I think Option 3 ( not sure though )

QUESTION 28.

It will be 4.9 ms^-2

The Value of g' is 9.8 ms^-2. When the distance will be doubled the Value will get Halved.

____________________________________

Answer 2

$\bf{Answer}$

$\red{Q.26}$What is the velocity of a body of mass 100 g having K.E. of 20 J ?

$<i>solution<i>$-

K.E=20j

1/2mv^2=20

mv^2=40

v^2=400. °•°[ mass,m=100g=.1kg] •°•v=20m/s

Momentum,p =mv=.1×20=2

$\red{Q.27}$As the frequency of a source decreases in a given medium, the wavelength of a periodic longitudinal wave

solution-
the speed of propagation depends upon the wavelength of the wave, and wavelength changes as the frequency changes. the speed of propagation depends only on the wavelength of the wave.the speed of propagation is constant in a given medium; only the wavelength changes as the frequency changes.

$\red{Q.28}$. What will be the value of acceleration due to gravity at a height 2R from the surface of earth?

Description

The acceleration of a body due to the attraction of the earth (radius R) at a distance 2R from the surface of the earth is (g = acceleration due to gravity at the surface of the earth)

g'g=(R/R+h)^2

=(R/R+2R)^2=1/9

∴g'=g°/9