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A wire has a mass (0.3±0.003)g, radius (0.5±0.005)mm and length (6±0.06)cm. The maximum percentage error in the measurement of its density is -
A). 1
B). 2
C). 3
D). 4

Answer
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Hint: In order to deal with this question we will use the formula of density which is the ratio of the mass of wire to the mass of volume, then we will proceed further by writing it in the form of error percentage.

Formula used: V=πr2L,ρ=mV=mπr2L,logab=logalogb

Complete step-by-step solution:
Given that
m+Δm=(0.3±0.003)gr+Δr=(0.5±0.005)mmL+ΔL=(6±0.06)cm
We know that the volume of the wire is given by, V=πr2L
Thus, the density of the wire, ρ=mV=mπr2L
Taking log on both the sides, we get
ρ=mπr2Llog(ρ)=log(mπr2L)log(ρ)=log(m)log(πr2L) [logab=logalogb]
Now after differentiating both the side, we get:
1ρdρ=1mdm+21rdr+1LdLΔρρ=Δmm+2Δrr+ΔLLΔρρ×100=Δmm×100+2Δrr×100+ΔLL×100
Notice that we had a negative sign which changed to positive after differentiating , this is because we are trying to compute the error in the measurement or the deviation of the result so it is cumulative i.e. irrespective of sign it will be positive.
Now, Let us substitute all the values known to us in the above-found formula, we get
Δρρ×100=Δmm×100+2Δrr×100+ΔLL×100Δρρ×100=0.0030.3×100+20.0050.5×100+0.066×100Δρρ×100=1+2+1Δρρ×100=4%
Hence, the maximum percentage error in the measurement of its density is 4%
So, the correct answer is option D.


Note: Error is the difference between the actual value of any physical quantity and its calculated value. In physics, there are essentially three types of mistakes, spontaneous mistakes, blunders, and systematic errors. In some data, the approximation error occurs between an exact value of the quantity under consideration and the approximation to it. An approximation error may occur because the data measured by the instrument is not accurate Or we use approximations instead of the real details.
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