You visited us 10 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Variation of G Due to Height and Depth

If the length...

Question

If the length of a simple pendulum is increased by $2$ %, then the time period

Increases by

2

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

Decreases by

2

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

Increases by

1

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

Decreases by

1

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

Open in App

Solution

The correct option is B Increases by $1$ %
Given that $l_{2} = 1.02 l_{1}$
We know that $T = 2 π \sqrt{(\frac{l}{g})}$
or $T \propto \sqrt{l}$
$∴ \frac{T_{2}}{T_{1}} = \sqrt{(\frac{l_{2}}{l_{1}})} = \sqrt{\frac{(1.02 l_{1})}{l_{1}}} = 1.01$
Thus time period increases by $1$ %

Suggest Corrections

Similar questions

Q. The time period of a simple pendulum is T. When the length is increased by 10 cm, the period is

T_{1}

.When the length is decreased by 10 cm, its periodis T

_{2}

. Then relation between T, T

_{1}

, T

_{2}

By how much will the potential of half cell cu+2/Cu change, if the solution is diluted to 100 times at 298 K?

(1) Increases by 59 mV

(2) Decreases by 59 mV

(3) Increases by 29.5 mv

(4) Decreases by 29.5 mV

If a simple pendulum is taken to a place where g decreases by 2%, then the time period

Q. Time period

T

of a simple pendulum of length

l

is given by

T = 2 π \sqrt{\frac{l}{g}} .

If the length is increased by

2 %

, then an approximate change in the time period is

When the length and width of a rectangle are increased by 5 cm each, its area increases by 100 cm². However, if the length of the rectangle is increased by 1 cm and its breadth is decreased by 1 cm, then the area of the rectangle decreases by 6 cm².

What is the original area of the rectangle?

Join BYJU'S Learning Program

If the length of a simple pendulum is increased by 2%, then the time period

The correct option is B Increases by 1%Given that l2=1.02l1We know that T=2π√(lg) or T∝√l∴T2T1=√(l2l1)=√(1.02l1)l1=1.01Thus time period increases by 1%

If the length of a simple pendulum is increased by $2$ %, then the time period

The correct option is B Increases by $1$ %
Given that $l_{2} = 1.02 l_{1}$
We know that $T = 2 π \sqrt{(\frac{l}{g})}$
or $T \propto \sqrt{l}$
$∴ \frac{T_{2}}{T_{1}} = \sqrt{(\frac{l_{2}}{l_{1}})} = \sqrt{\frac{(1.02 l_{1})}{l_{1}}} = 1.01$
Thus time period increases by $1$ %