a person have a gold coin of diameter 2cm and thickess 2mm to a goldsmith for melting and drawing into a wire of diameter 1mm . the gold smith gave back a wire of gold of length 76cm with diameter 1mm. if the wastage allowed in melting could be a maximum of 5% of the total volume , question is whether gold smith was honest or not
The volume of the coin is given by \[V _{c}=\frac{\pi d ^{2}}{4}\times x\] where d is the diameter and x is the thickness. The volume of the wire is given by \[V _{w}=\frac{\pi d ^{2}}{4}\times y\] where d is the diameter and y is the length of the wire. First you should convert the dimensions to mm in each case. Diameter of coin =20mm Length of wire = 760mm Next calculate the two values of volume and find the percentage reduction in volume of the wire against volume of the coin.
to find percentage in reduction={[V(wire)-V(coin)]*100}V(wire)
no, to find the percentage in reduction , it is given by this formula =V(wire)-V(coin)/V(coin)*100
k
[(y-x)/x]%
The volume of the coin is the reference amount in the calculation of the percentage reduction in volume. The percentage reduction in the volume of gold originally in the coin is required. Therefore the percentage reduction in volume is \[\frac{V _{c}-V _{w}}{V _{c}}\times 100\]
for any question if you want to find out the change in percentage then it it given by =(final value-initial value)/initial value *100
The reduction in volume of gold is \[\frac{\pi}{4}[(20^{2}\times 2)-760]=?\]
31.4
Using the value of pi in my calculator the reduction in volume = 31.4159 mm^3
yup you r right @kropot72 , me too getting the same answer
The volume of the coin is \[V _{c}=\frac{\pi \times 20^{2}}{2}=?\]
200pi
Correct. Percentage reduction in volume of gold originally in coin is \[\frac{31.4159}{200\times \pi}\times 100=?\]
5.0025
Not quite. The calculation gives the result 4.99999%
k
I get the result 4.99999% on both a Casio and a Sharp calculator.
yup u r right
Cool!
it means gold smith was honest
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