A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its cross-sectional area is 4.9×10-7 m2. If the mass is pulled a little in the vertically

Question

A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its cross-sectional area is 4.9×10-7 m2. If the mass is pulled a little in the vertically (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

ω=√(K/m)=√(YA/lm)=√((n×109)(4.9×10-7)/1×0.1) Putting, ω=140 rad s-1 in above equation we get, n=4

Q. A $0.1 k g$ mass is suspended from a wire of negligible mass. The length of the wire is $1 m$ and its cross-sectional area is $4.9 \times 1 0^{- 7} m^{2}$ . If the mass is pulled a little in the vertically

$ω = \frac{K}{m} = \frac{Y A}{l m} = \frac{( n \times 1 0 ^{9} ) ( 4.9 \times 1 0 ^{- 7} )}{1 \times 0.1}$
Putting, $ω = 140$ rad $s^{- 1}$ in above equation we get, n=4

Q. A 0.1kg mass is suspended from a wire of negligible mass. The length of the wire is 1m and its cross-sectional area is 4.9×10−7m2. If the mass is pulled a little in the vertically

ω=mK​​=lmYA​​=1×0.1(n×109)(4.9×10−7)​​ Putting, ω=140 rad s−1 in above equation we get, n=4

Q. A $0.1 k g$ mass is suspended from a wire of negligible mass. The length of the wire is $1 m$ and its cross-sectional area is $4.9 \times 1 0^{- 7} m^{2}$ . If the mass is pulled a little in the vertically

$ω = \frac{K}{m} = \frac{Y A}{l m} = \frac{( n \times 1 0 ^{9} ) ( 4.9 \times 1 0 ^{- 7} )}{1 \times 0.1}$
Putting, $ω = 140$ rad $s^{- 1}$ in above equation we get, n=4