Time is measured using a stop watch of least count 0.1 second In 10 oscilation time taken is 20.0 second Find maximum permissible error in time period .
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A radioactive element gets spilled over the floor of a room. Its half life period is 30 days. If its initial activity is ten times the permissible val
Correct Answer - A a) Let the initial activity for safe working = A `therefore` Initial activity `[A]_(0) = 10 A` `k=0.693/30` days `t=(2.303)/t log ([A]_(0))/([A])` `t=(2.303 xx (30"days"))/(0.693) log 10`...
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Determine the maximum permissible discount which a company can allow at the time of reissue of forfeited shares in the following cases: (i) A share of
(i) Money received in respect of capital on shares forfeited is Rs. 5. Therefore, maximum permissible discount to be allowed is also Rs. 5. (ii) Money received in respect of...
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In resonance tube exp we find `l_(1) = 25.0 cm` and `l_(2) = 75.0cm` The least count of the scale used to measure l is `0.1cm` If there is no error in
`V = 2f_(0) (l_(2) - l_(1))` `(dV) = 2f_(0) (dl_(2) - dl_(1))` `(DeltaV)_(max) = max of [2f_(0) (+- Deltal_(2) + Deltal_(1))` `Deltal_(1)=` least count of the scale `= 0.1 cm`...
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If measured value of resistance `R = 1.05Omega` wire diameter `d = 0.60 mm` and length `l = 75.3 cm` If maximum error in resistance measurment is `0.0
`((Deltarho)/(rho))_(max)=(DeltaR)/(R)+2(Deltad)/(d)+(Deltal)/(l)` `DeltaR = 0.01 Omega` `Deltad = 0.01 mm` (least count) `Deltal = 0.1 cm` (least count) `((Deltarho)/(rho))_(max)=((0.01Omega)/(1.05Omega)+2(0.01mm)/(0.60mm)+(0.1cm)/(75.3cm))xx100=4.3%` .
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If mass and specific heat capacity of calormeter is negligible what would be maximum permissible error is `S` Use the data mentioned below `m_(1) rarr
If `m_(1) rarr 0, S_(c) rarr0` `S_(l) = (Vit)/(m_(2)(theta_(f)-theta_(0))` `(DeltaS_(l))/(S_(l))=(DeltaV)/(V)+(DeltaI)/(I)+(Deltat)/(t)+(Deltam_(2))/(m_(2))+(Deltatheta_(f)+Deltatheta_(0))/(theta_(f)-theta_(0))=(0.1)/(10.0)+(0.01xx10^(2))/(1.00xx10^(2))+(0.01)/(1.00)+(1+1)/(50)=8%` .
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In the previous question, find the maximum permissible error in resistivity and resistance.
`((DeltaR)/(R))_(max)=(Deltai)/(i)+(Deltav)/(v)=(0.1)/(10.0)+(0.1)/(100.0)=1.1%rArr((dp)/(rho))_(max)=2.42%` .
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In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate
Let at time `t`, the number of bacteria be `y` then `(dy)/(dt)propyimplies (dy)/(dt)=ky`, where `k` is proportional constant. `implies (dy)/(y)=k dt` `impliesint(dy)/(y)=intk dt` `implies log|y|=kt+C` .........`(1)` Initially at `t=0`, `y=100000`...
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A paricle performs `S.H.M.` with time period `T`. The time taken by the particle to move from hall the amplitude to the maximum dispalcement is `T/2`
Correct Answer - C Required time `= T/4 - T/12 = T/6`
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The mass of a cube has error of 2% and the length of its edge has error of 1%. What will be the maximum percentage error in calculation of its density
Density of cube `(p)=(M)/(V)=(M)/L^(3)) therefore((Deltap)/(P))_("max")xx100=((DeltaM)/(M)xx100)+3(DeltaL)/(L)xx100)=2%+3(1%)=5%`
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A survey shows that 63 % of the pepole watch a news channel whereas , 76 % watch an entertiament channel at a particular time If X% of the pepole watc
Correct Answer - C Let A be the set of those pepole who watch news channel and B be the set those people who watch entertainment channel then c(A)= 63...
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