Question

Breaking tension of the string is 60 N.A body of mass 0.3kg is attached to it and rotate in a circle of radius 1m what will be minimum period of revolution before string breaksv​

Answer 1

Answer:

10m/s

Explanation:

TheratiotothemamimumT

2

totheminimumtensionT

1

is−

T

1

T

2

=4

T

2

=4T

1

Nowthedifferencebetweenthetwo6mg;

Hencewehave

T

2

−T

1

=6mg

∴4T

1

−T

1

=6mg

3T

1

=6mg

T

1

=2mg

Nowtensionatthetopofthecircleis−

T

1+mg

=

r

mv

1

2

=

L

mv

1

2

2mg+mg=

3

10

mv

1

2

v

1

2

=3g

3

10

=10g

v

1

=

10g

=

10×10

=10m/s

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