Potential Energy Formulae
Trending Questions
(Assume that potential energy at mean position is zero)
- √E21+E22
- E1+E2+2√E1E2
- √E1E2
- E1+E2
Block A of mass m is performing SHM of amplitude a. Another block B of mass m is gently placed on A when it passes through mean position and B sticks to A. Find the amplitude of new S.H.M.
- a√2
- a2
- a2√2
- a√2
- 100 Newton/m
- 1000 Newton-m
- 1000 Newton/m
- 1000 watts
If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2√π cm then the Reynolds number for the flow is (density of water = 103kg/m3 and viscosity of water = 10−3 Pa.s) approximately
- 5500
- 1100
- 11, 000
- 550
- A√2
- A2
- √3A2
- Zero
Column IColumn II(A) Point J is position of(p) Neutral equilibrium(B) Point K is position of(q) Unstable equilibrium(C) Point L is position of(r) Stable equilibrium(D) Point M is position of(s) No equilibrium
- A→q;B→s;C→r;D→p
- A→r;B→q;C→s;D→p
- A→s;B→q;C→r;D→p
- A→s;B→p;C→r;D→q
- 1×102 N/m
- 1.5×102 N/m
- 2×102 N/m
- 3×102 N/m
A particle oscillates about the equilibrium position x0 subject to a force that has an associated potential energy U(x). Which of the following statements about U(x) is/are true:
u(x) must be symmetric about x0
u(x) must have a minimum at x0
u(x) must be positive in the vicinity of x0
u(x) may have a minimum at x0
- Maximum K.E. is 160 J
- Maximum potential energy is 100 J
- Maximum P.E. is 160 J
- Minimum P.E. is zero
- 50 J
- 200 J
- 100 J
- 150 J
- 50 J
- 250 J
- 100 J
- 200 J
- 2π5 s
- 2√2π5 s
- √2π5 s
- 4π5 s
- 100N.m−1
- 200N.m−1
- 50N.m−1
- 500N.m−1
Column IColumn II(A) Kinetic energy(p) Half the maximum value(B) Potential energy(q) 3/4 times the maximum value(C) Acceleration(r) 1/4 times the maximum value(s) Cannot say anything
- A→q, B→s, C→p
- A→q, B→p, C→s
- A→q, B→p, C→s
- A→q, B→s, C→s
- two
- three
- four
- 3K2
- 2K5
- K
- 5K2
- WP=WQ ; WP>WQ
- WP=WQ ; WP=WQ
- WP>WQ ; WQ>WP
- WP<WQ ; WQ<WP
- WP=WQ ; WP>WQ
- WP=WQ ; WP=WQ
- WP<WQ ; WQ<WP
- WP>WQ ; WQ>WP
During a time-interval of exactly one period of vibration of a tuning fork, the emitted sound travels a distance
equal to the length of tuning fork
of about
which decrease with time
of one wavelength in air
- 1×102 Nm−1
- 1.5×102 Nm−1
- 2×102 Nm−1
- 3×102 Nm−1
A force shown in the F−x graph is applied to a 5 kg cart, which then climbs up a ramp as shown. The maximum height Ymax the cart can reach (in m) is
Plancks constant has the dimensions as that of
Power
Energy
Angular Momentum
Linear Momentum
- √Gma
- √2Gma
- √Gm2a
- √2Gm3a
Column IColumn II(A) Kinetic energy(p) Half the maximum value(B) Potential energy(q) 3/4 times the maximum value(C) Acceleration(r) 1/4 times the maximum value(s) Cannot say anything
- A→q, B→p, C→s
- A→q, B→s, C→p
- A→q, B→p, C→s
- A→q, B→s, C→s
Why should Gibbs free energy be negative?
Reason : Potential energy is always negative and if it is greater than kinetic energy total mechanical energy will be negative.
- Both assertion and reason are true but reason is not the correct explanation of assertion.
- Both assertion and reason are true and reason is the correct explanation of assertion.
- Assertion is true but reason is not true.
- Both assertion and reason are not true.
- 2A
- 3A
- 4A
- A
- K0
- zero
- K0/2
- not obtainable