The time period of a second pendulum has been increased. To make its time period 2s, its length will be.

The time period of a second pendulum has been increased, To make its time period 2s, its length will be:

A

Increased

B

Decreased

C

Nothing to be done

D

Everyone is correct

The periodic time of one oscillation for a simple pendulum is (where $$l$$ = Length of the pendulum.)

A

$$\frac{1}{{2\pi }}\sqrt {\frac{l}{{\text{g}}}} $$

B

$$\frac{1}{{2\pi }}\sqrt {\frac{{\text{g}}}{l}} $$

C

$$2\pi \sqrt {\frac{l}{{\text{g}}}} $$

D

None of these

The equivalent length of a simple pendulum which gives the same frequency as compound pendulum is

A

$$\frac{{\text{h}}}{{{\text{k}}_{\text{G}}^2 + {{\text{h}}^2}}}$$

B

$$\frac{{{\text{k}}_{\text{G}}^2 + {{\text{h}}^2}}}{{\text{h}}}$$

C

$$\frac{{{{\text{h}}^2}}}{{{\text{k}}_{\text{G}}^2 + {{\text{h}}^2}}}$$

D

$$\frac{{{\text{k}}_{\text{G}}^2 + {{\text{h}}^2}}}{{{{\text{h}}^2}}}$$

The length of a simple pendulum which gives the same frequency as the compound pendulum, is

A

$${\text{kG}} + {l_1}$$

B

$${\text{k}}{{\text{G}}^2} + {l_1}$$

C

$$\frac{{{\text{k}}{{\text{G}}^2} + l_1^2}}{{{l_1}}}$$

D

$$\frac{{{\text{kG}} + l_1^2}}{{{l_1}}}$$

The equivalent length of a simple pendulum which gives the same frequency as a compound pendulum is

A

$$\frac{{\text{h}}}{{{\text{k}}_{\text{G}}^2 + {{\text{h}}^2}}}$$

B

$$\frac{{{\text{k}}_{\text{G}}^2 + {{\text{h}}^2}}}{{\text{h}}}$$

C

$$\frac{{{{\text{h}}^2}}}{{{\text{k}}_{\text{G}}^2 + {{\text{h}}^2}}}$$

D

$$\frac{{{\text{k}}_{\text{G}}^2 + {{\text{h}}^2}}}{{{{\text{h}}^2}}}$$

Time period and length of a second’s pendulum respectively are

A

1 sec and 99.4 cm

B

1 sec and 92.7 cm

C

2 sec and 99.4 cm

D

2 sec and 92.7 cm

The length of each of square A is increased by 100 percent to make square B.Again the length of the each side of square B is increased by 50 peacent to make square C.By of the area df square A and B?

A

75%

B

80%

C

85%

D

90%

The length of each side of square A is increased by 100 percent to make square B. Again the length of each side of square B is increased by 50 percent to make square C. By what percent is the area of square C greater than the sum of the areas of square A and B?

A

75%

B

80%

C

85%

D

90%

The length of a rectangle is twice its breadth . If its length is decreased y 5 cm and breadth is increased by 5 cm , the area of the rectangle is increased by 75 sq. cm . Find the length of the rectangle.

A

20 cm

B

25 cm

C

30 cm

D

Try your self

Two men and women are entrusted with a task. The second man needs three hours more to cope up with the job than the second man and the woman would need working together. The first man, working alone, would need as much time as second man and the woman working together. The first man working alone, would spend eight hours less than the double period of the time second man would spend working alone. How much time would the two men and the women need to complete the task if they all asked together?

A

1 hour

B

2 hours

C

3 hours

D

4 hours

E

5 hours

The time period of a second pendulum has been increased. To make its time period 2s, its length will be.

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