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`        if the error in the measurement of momentum of a particle is 10% and mass is known exactly,the permissible error in the determination of kinetic energy is `
4 years ago

```							2*10= 20%
```
4 years ago
```							are u sure??the answer in my book is 21%
```
4 years ago
```							yes 21%u can apply calucus only when error is very small or less then 1 here error is 10% so by calucus u get 2*10 =20 which is wrong to do such problems first take P1=100 and calulate KE1now put p2=110 and calculate KE2 then find the errorPLEASE APPROVE
```
4 years ago
```							i did not get it i am extemely sorry.i mean why should we put p=110
```
4 years ago
```							Here ,           K.E- 1/2mv^2Mass is known exactly but there is always a minimum error in any calculation so, Here we let the minimum in mass =1% Now,      ∆K/K=∆ M/M +2∆V/V                 = 1% + 2(10%)                 = 1% + 20%                 = 21%
```
2 years ago
```							P=mv
Ke = 1/2 mv^2
Ke = P^2/2m
Here error is 10%
Thus
P'= 10% of P
P'=0.1P
Ke'=P'/2m
=1.21×Ke
Percentage error = (Final - Initial)×100/initial
=(1.21Ke - Ke)×100/Ke
= 21%

```
11 months ago
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