CBSE 11TH PHYSICS - Online Test

W=Pt=10×104×2.5×60=150×105J

Energy used in heating = 50% =75 x 105J

Q=msΔTΔT=Qms

ΔT=75×1058000×0.91

ΔT=103∘C

There are three rules on determining how many significant figures are in a number:

• Non-zero digits are always significant.
• Any zeros between two significant digits are significant.
• A final zero or trailing zeros in the decimal portion ONLY are significant.

So keeping these rules in mind, there are 4 significant digit

Assuming bigger pendulum as the pendulum of bigger time period i.e.having time period 5T/4

As the two pendulums start at the same time in simple harmonic motion from the mean position,their initial phase angles are zero

The bigger pendulum completes its one oscillation in time t=5T/4 and comes back to the mean position.During this time the point on reference circle associated with SHM rotates by an angle 2π

If ω represents the angular velocity of the reference point associated with SHM of first pendulum of time period T,then the phase of the this pendulum after 5T/4 s willbe

ω5T/4=ωT+ωT/4

=2π+2π/4

2π+π/2

Hence the phase difference

2π+π/2−2π=π/2

= 900

Initial velocity u = 30 m/s

As it stops then final velocity v = 0 m/s

Time taken t = 2.0 s

We know,

v-u = at

=> 0-30= 2a

=> a = −302=−15 m/s2

-ve sign shows velocity is decreasing.

We know that at the maximum height, the velocity of the ball is 0 m/s.

We also know that the ball takes the same time to reach its maximum height as it takes to travel from its maximum height to the initial position, due to time symmetry. The time taken is half the total time.

Therefore, we have the following information for the second (downward) part of the motion of the ball:

t = 5 second; half the total time

vtop=0 m/s

g = 9.8 m/s​​​​​​2 downward

s = ?

As we know

s = ut+12gt2

Put all the given value,

=> s = 0×5+12×9.8×25

=> s = 122.5 m

When the lift falls freely under gravity, a=g

Therefore Apparent weight, 

R=m(g−a)=m(g−g)=0

This is the condition of weightlessness.