A sphere of mass 0.20 kg is attached to an inextensible string of length 0.5m whose upper end is fixed to the ceiling. The sphere is made to describe a horizontal circle of radius 0.3 m. The speed of the sphere will be

Question

A sphere of mass 0.20 kg is attached to an inextensible string of length 0.5m whose upper end is fixed to the ceiling. The sphere is made to describe a horizontal circle of radius 0.3 m. The speed of the sphere will be (A) 1.5 m/s (B) 2.5 m/s (C) 3.2 m/s (D) 4.7 m/s

Tardigrade Admin · Accepted Answer

As shown in figure T sin θ =(mv2/r) T cos θ =mg So, tan θ =(v2/rg)=(r/√l2-r2) ∴ v=[ (r2g/(l2-r2)1/2) ]1/2 =[ (0.09× 10/(0.25-0.09)1/2) ]1/2=1.5 m/s

Q. A sphere of mass 0.20 kg is attached to an inextensible string of length 0.5m whose upper end is fixed to the ceiling. The sphere is made to describe a horizontal circle of radius 0.3 m. The speed of the sphere will be

1.5 m/s

2.5 m/s

3.2 m/s

4.7 m/s

As shown in figure
$T sin θ = \frac{m v ^{2}}{r}$ $T cos θ = m g$ So, $tan θ = \frac{v ^{2}}{r g} = \frac{r}{l ^{2} - r ^{2}}$ $∴$ $v = [\frac{r ^{2} g}{( l ^{2} - r ^{2} ) ^{1/2}}]^{1/2}$ $= [\frac{0.09 \times 10}{( 0.25 - 0.09 ) ^{1/2}}]^{1/2} = 1.5 m / s$

Q. A sphere of mass 0.20 kg is attached to an inextensible string of length 0.5m whose upper end is fixed to the ceiling. The sphere is made to describe a horizontal circle of radius 0.3 m. The speed of the sphere will be

1.5 m/s

2.5 m/s

3.2 m/s

4.7 m/s

As shown in figure Tsinθ=rmv2​ Tcosθ=mg So, tanθ=rgv2​=l2−r2​r​ ∴ v=[(l2−r2)1/2r2g​]1/2 =[(0.25−0.09)1/20.09×10​]1/2=1.5m/s

As shown in figure
$T sin θ = \frac{m v ^{2}}{r}$ $T cos θ = m g$ So, $tan θ = \frac{v ^{2}}{r g} = \frac{r}{l ^{2} - r ^{2}}$ $∴$ $v = [\frac{r ^{2} g}{( l ^{2} - r ^{2} ) ^{1/2}}]^{1/2}$ $= [\frac{0.09 \times 10}{( 0.25 - 0.09 ) ^{1/2}}]^{1/2} = 1.5 m / s$