An insect crawls up a hemispherical surface very slowly (see the figure). The coefficient of friction between the surface and the insect is 1/3. If the line joining the centre of the hemispherical surface to the insect makes an angle a with the vertical, the maximum possible value of α is given

Question

An insect crawls up a hemispherical surface very slowly (see the figure). The coefficient of friction between the surface and the insect is 1/3. If the line joining the centre of the hemispherical surface to the insect makes an angle a with the vertical, the maximum possible value of α is given (A)  cot α = 3 (B)  tan α = 3 (C)  sec α = 3 (D) cosec α = 3

Tardigrade Admin · Accepted Answer

Equilibrium of insect gives N = mg cos α ...(i) μ N = mg sin α ...(i) From Eqs. (i) and (ii), we get cot α=(1/μ)=3

$cot α = 3$

$tan α = 3$

$sec α = 3$

$cosec α = 3$

Equilibrium of insect gives
$N = m g cos α ... (i)$
$μ N = m g sin α ... (i)$
From Eqs. (i) and (ii), we get
$cot α = \frac{1}{μ} = 3$

Q. An insect crawls up a hemispherical surface very slowly (see the figure). The coefficient of friction between the surface and the insect is 1/3. If the line joining the centre of the hemispherical surface to the insect makes an angle a with the vertical, the maximum possible value of α is given

cotα=3

tanα=3

secα=3

cosecα=3

Equilibrium of insect gives N=mgcosα...(i) μN=mgsinα...(i) From Eqs. (i) and (ii), we get cotα=μ1​=3

$cot α = 3$

$tan α = 3$

$sec α = 3$

$cosec α = 3$

Equilibrium of insect gives
$N = m g cos α ... (i)$
$μ N = m g sin α ... (i)$
From Eqs. (i) and (ii), we get
$cot α = \frac{1}{μ} = 3$