A) \[\sigma \]
B) \[{{V}_{A}},{{V}_{B}}\]
C) \[{{V}_{C}}\]
D) \[{{V}_{C}}={{V}_{A}}\ne {{V}_{B}}\]
Correct Answer: C
Solution :
Key Idea If the particle is moving in a straight line under the action of a constant force then distance covered \[{{v}_{S}}=v+{{v}_{B}}\] Since the body start from rest u = 0 \[=\frac{1000}{100}=10\,\text{m}{{\text{s}}^{-1}}\]\[{{v}_{s}}=10+10=20m{{s}^{-}}^{1}\] Now, \[T=\frac{2\pi r}{v}=\frac{2\pi }{v}\times \frac{mv}{Bq}\] ..(i) and \[T=\frac{2\pi m}{Bq}\] ...(ii) Dividing Eq. (i) and Eq. (ii), we get \[v=128\,\text{m}{{\text{s}}^{-1}}\]\[5\lambda =4\]\[k=\frac{2\pi }{\lambda }=\frac{2\pi \times 5}{4}=7.85\]
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