A metallic disc is being heated. Its area (in `m^(2)`) at any time t (in sec) is given by `A=5t^(2)+4t`. Calculate the rate of increase in area at `t=3sec`.
Answered Feb 05, 2023
Correct Answer - `34m^(2)//sec`
`y=sqrt((1-sin 2x)/(1+ sin 2x))=(cos x - sin x)/(cos x + sin x)=(1-tan x)/(1+ tan x)=tan ((pi)/(4)-x)` `therefore (dy)/(dx)=-sec^(2)((pi)/(4)-x)`
Correct Answer - `[(piR^(3))/(2mu_(0))]`
1. Interest rate 2. Expansion of industrial sector
Correct Answer - `10 pi cm//sec`
Correct Answer - `(a) 2 cm//sec (b) 4 cm^(2)//sec`,Increase
Correct Answer - (i) 10 sec (ii) 50 m//s (iii) 125 m (iv) 80 m
Correct Answer - `a=(10)/(19)m//s^(2)`
Correct Answer - `(dP)/(dt) = 0.8 cm//s`
Given: V0 = 2 × 10-4 m3, T0 = 0°C,T = 80°C, γr = 4 × 10-4 K-1, T – T0 = 80 – 0 = 80°C To find: Increase in volume (∆V) Formula: ∆V = V0γr(T – T0) Calculation: From formula. ∆V =...
Correct option is: (B) 0.32 m2
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