You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Buoyant Force

Two springs A...

Question

Two springs A and B ( $k_{A} = 2 k_{B}$ ) are stretched by applying forces of equal magnitudes at the four ends. If the energy stored in A is E, then in B is (assume equilibrium):

E / 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 E

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

E

No worries! We‘ve got your back. Try BYJU‘S free classes today!

E / 4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $2 E$
Force is same, thus $0.5 k_{a} x_{a} = 0.5 (k_{a} / 2) (x_{b})$

Thus $x_{b} = 2 x_{a}$

Energy in $A = U = 0.5 k_{A} x^{2} = E$

Energy in $B = 0.5 k_{b} x^{2} = 0.5 (k_{A} / 2) (2 x)^{2} = 2 E$

Suggest Corrections

Similar questions

Q. Two springs

A

and

B

(

k_{A} = 2 k_{B}

) are stretched by applying forces of equal magnitudes at the four ends. If the energy stored in

A

E

, that in

B

Two springs A and B ( $k_{A}$ = 2 $k_{B}$ ) are stretched by applying forces of equal magnitudes at the four ends. If the energy stored in A is E, that in B is

Q. Two springs A and B having spring constant

K_{A}

and

K_{B}

(

K_{A}

2 K_{B}

) are stretched by applying force of equal magnitude. If energy stored in spring A is E then energy stored in B will be :

Q. Two springs

A

and

B

are stretched by applying forces of equal magnitudes. If the energy stored in

A

is E, then the energy stored in B will be-
(Given relation between spring constants of

A

and

B

(k_{A} = 2 k_{B})

)

Q. Two springs

P

and

Q

of force constants

k_{P}

and

k_{Q} (k_{Q} = \frac{k_{P}}{2})

are stretched by applying forces of equal magnitude. If the energy stored in

Q

E

, then the energy stored in

P

Join BYJU'S Learning Program

Two springs A and B (kA=2kB) are stretched by applying forces of equal magnitudes at the four ends. If the energy stored in A is E, then in B is (assume equilibrium):

The correct option is B 2EForce is same, thus 0.5kaxa=0.5(ka/2)(xb)Thus xb=2xaEnergy in A=U=0.5kAx2=EEnergy in B=0.5kbx2=0.5(kA/2)(2x)2=2E

Two springs A and B ( $k_{A} = 2 k_{B}$ ) are stretched by applying forces of equal magnitudes at the four ends. If the energy stored in A is E, then in B is (assume equilibrium):

The correct option is B $2 E$
Force is same, thus $0.5 k_{a} x_{a} = 0.5 (k_{a} / 2) (x_{b})$

Thus $x_{b} = 2 x_{a}$

Energy in $A = U = 0.5 k_{A} x^{2} = E$

Energy in $B = 0.5 k_{b} x^{2} = 0.5 (k_{A} / 2) (2 x)^{2} = 2 E$