An insect is at the bottom of a hemispherical ditch of radius 1 m. It crawls up the ditch but starts slipping after it is at height h from the bottom. If the coefficient of friction between the ground and the insect is 0.75, then h is: (g=10 m s-2)

Question

An insect is at the bottom of a hemispherical ditch of radius 1 m. It crawls up the ditch but starts slipping after it is at height h from the bottom. If the coefficient of friction between the ground and the insect is 0.75, then h is: (g=10 m s-2) (A) 0.80 m (B) 0.60 m (C) 0.45 m (D) 0.20 m

Tardigrade Admin · Accepted Answer

For balancing operatornamemg sin θ= f mg sin θ=μ mg cos θ tan θ=μ tan θ=(3/4)

h = R - R cos θ =R-R((4/5))=(R/5) h =( R /5)=0.2 m

Q. An insect is at the bottom of a hemispherical ditch of radius $1 m$ . It crawls up the ditch but starts slipping after it is at height $h$ from the bottom. If the coefficient of friction between the ground and the insect is $0.75,$ then $h$ is : $(g = 10 m s^{- 2})$

$0.80 m$

$0.60 m$

$0.45 m$

$0.20 m$

For balancing $mg sin θ = f$
$m g sin θ = μ m g cos θ$
$tan θ = μ$
$tan θ = \frac{3}{4}$

$h = R - R cos θ$
$= R - R (\frac{4}{5}) = \frac{R}{5}$
$h = \frac{R}{5} = 0.2 m$

Q. An insect is at the bottom of a hemispherical ditch of radius 1m. It crawls up the ditch but starts slipping after it is at height h from the bottom. If the coefficient of friction between the ground and the insect is 0.75, then h is : (g=10ms−2)

0.80m

0.60m

0.45m

0.20m