Tag: vernier calliper and screw gauge
Questions Related to vernier calliper and screw gauge
Obs, No. | Voltage/V | Current/mA | Obs No. | Voltage/V | Current/mA |
---|---|---|---|---|---|
1 | 1.0 | 40 | 4 | 7.0 | 160 |
2 | 3.0 | 80 | 5 | 9.0 | 200 |
3 | 5.0 | 120 |
A physical quantity $Q$ is released to four observable $x,y,z$ and $t$ by the relation
$Q=\dfrac {x^{2/5}z^{3}}{y\sqrt {t}}$
The percentage errors of measurement in $x,y,z$ and $t$ are $2.5\%,2\%,0.5\%$ and $1\%$ receptively. The percentage error in $Q$ will be
A steel tape is calibrated at $20 ^ { \circ } \mathrm { C }$. On a cold day when the temperature is $- 15 ^ { \circ } \mathrm { C }$ percentage error in the tape will be $\left[ \alpha _ { \text { steel } } = 1.2 \times 10 ^ { - 5 \cdot 0 } C ^ { - 1 } \right]$
The random error in the arithmetic mean of 100 observations is x, then random error in the arithmetic mean of 400 observation would be
A faulty thermometer has its fixed point marked 5${ \circ } _{ C }$ and 95${ \circ } _{ C }$. this thermometer reads the temperature of a body as ${ 59 }^{ \circ }$. The correct temperature on Celsius scale is
If the error in measurement of momentum of a particle is $10$% and mass is known exactly,the permissible error in the determination of kinetic energy is
Length of a thin cylinder as measured by vernier callipers having least cound $0.01\ cm$ is $3.25\ cm$ and its radius of cross-section is measured by a screw gauge having least count $0.01\ mm$ as $2.75\ mm$. The percentage error in the measurement of volume of the cylinder will be
If the error in the measurement of radius of a sphere is 1%, then the error in the measurement of volume will be :
The speed of a body is measured with a positive error of 30%. If the mass of the body is known exactly, the kinetic energy is calculated with an error of
While measuring the acceleration due to gravity by a simple pendulum, a student makes a positive error of 1% in the length of the pendulum and a negative error of 3% in the value of time period. His per-centre error in the n=measurement of g by the relation $g={ 4\pi }^{ 2 }\left( I/T^{ 2 } \right) $ will be