Answer:
D) -Q
Explanation:
The charge inserted will induce -Q charge on the inner surface and + Q on the outer surface of the shell . This charge is called bound charge because it remained attached with opposite charge inserted inside.
Answer:
Explanation:
In case of gas , work done
W = ∫ p dV , p is pressure and dV is small change in volume
the limit of integration is from Vi to Vf .
= ∫ p dV
= ∫ p₀
dV
= p₀
/ (
)
= - 5p₀ 
Taking limit from Vi to Vf
W = - 5 p₀ (
) ltr- atm.
Answer: 35*10^3 N/m
Explanation: In order to explain this problem we know that the potential energy for spring is given by:
Up=1/2*k*x^2 where k is the spring constant and x is the streching or compresion position from the equilibrium point for the spring.
We also know that with additional streching of 2 cm of teh spring, the potential energy is 18J. Then it applied another additional streching of 2 cm and the energy is 25J.
Then the difference of energy for both cases is 7 J so:
ΔUp= 1/2*k* (0.02)^2 then
k=2*7/(0.02)^2=35000 N/m
<h2>Answer: at an angle

below the inclined plane.
</h2>
If we draw the <u>Free Body Diagram</u> for this situation (figure attached), taking into account only the gravity force in this case, we will see the weight
of the block, which is directly proportional to the gravity acceleration
:

This force is directed vertically at an angle
below the inclined plane, this means it has an X-component and a Y-component:



Therefore the correct option is c
Answer:
Explanation:
Two frequencies with magnitude 150 Hz and 750 Hz are given
For Pipe open at both sides
fundamental frequency is 150 Hz as it is smaller
frequency of pipe is given by

where L=length of Pipe
v=velocity of sound
for n=1
and f=750 is for n=5
thus there are three resonance frequencies for n=2,3 and 4
For Pipe closed at one end
frequency is given by

for n=0


for n=2

Thus there is one additional resonance corresponding to n=1 , between
and 