Write difference between Mistake and Error.
Write difference between Mistake and Error.
| mistake | error |
|
$(1)$ mistake iws caused due to carelessness wrong calculation , improper method of measurment. |
$(1)$ Error is caused due to imperfect design of measuring instrument , observation limitation of individuals |
| $(2)$ by taking proper care the mistake can be sort out completely. | $(2)$ by taking more observation by using instrument having smaller least copunt error can be eliminated |
Similar Questions
Durring Searle's experiment, zero of the Vernier scale lies between $3.20 \times 10^{-2} m$ and $3.25 \times 10^{-2} m$ of the main scale. The $20^{\text {th }}$ division of the Vernier scale exactly coincides with one of the main scale divisions. When an additional load of $2 \ kg$ is applied to the wire, the zero of the Vernier scale still lies between $3.20 \times 10^{-2} m$ and $3.25 \times 10^{-2} m$ of the main scale but now the $45^{\text {th }}$ division of Vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is $2 m$. and its cross-sectional area is $8 \times 10^{-7} m ^2$. The least count of the Vernier scale is $1.0 \times 10^{-5} m$. The maximum percentage error in the Young's modulus of the wire is
Durring Searle's experiment, zero of the Vernier scale lies between $3.20 \times 10^{-2} m$ and $3.25 \times 10^{-2} m$ of the main scale. The $20^{\text {th }}$ division of the Vernier scale exactly coincides with one of the main scale divisions. When an additional load of $2 \ kg$ is applied to the wire, the zero of the Vernier scale still lies between $3.20 \times 10^{-2} m$ and $3.25 \times 10^{-2} m$ of the main scale but now the $45^{\text {th }}$ division of Vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is $2 m$. and its cross-sectional area is $8 \times 10^{-7} m ^2$. The least count of the Vernier scale is $1.0 \times 10^{-5} m$. The maximum percentage error in the Young's modulus of the wire is
- [IIT 2014]
A student uses a simple pendulum of exactly $1 \mathrm{~m}$ length to determine $\mathrm{g}$, the acceleration due to gravity. He uses a stop watch with the least count of $1 \mathrm{sec}$ for this and records $40$ seconds for $20$ oscillations. For this observation, which of the following statement$(s)$ is (are) true?
$(A)$ Error $\Delta T$ in measuring $T$, the time period, is $0.05$ seconds
$(B)$ Error $\Delta \mathrm{T}$ in measuring $\mathrm{T}$, the time period, is $1$ second
$(C)$ Percentage error in the determination of $g$ is $5 \%$
$(D)$ Percentage error in the determination of $g$ is $2.5 \%$
A student uses a simple pendulum of exactly $1 \mathrm{~m}$ length to determine $\mathrm{g}$, the acceleration due to gravity. He uses a stop watch with the least count of $1 \mathrm{sec}$ for this and records $40$ seconds for $20$ oscillations. For this observation, which of the following statement$(s)$ is (are) true?
$(A)$ Error $\Delta T$ in measuring $T$, the time period, is $0.05$ seconds
$(B)$ Error $\Delta \mathrm{T}$ in measuring $\mathrm{T}$, the time period, is $1$ second
$(C)$ Percentage error in the determination of $g$ is $5 \%$
$(D)$ Percentage error in the determination of $g$ is $2.5 \%$
- [IIT 2010]
A metal wire has mass $(0.4 \pm 0.002)\,g$, radius $(0.3 \pm 0.001)\,mm$ and length $(5 \pm 0.02) \,cm$. The maximum possible percentage error in the measurement of density will nearly be $.......\%$
A metal wire has mass $(0.4 \pm 0.002)\,g$, radius $(0.3 \pm 0.001)\,mm$ and length $(5 \pm 0.02) \,cm$. The maximum possible percentage error in the measurement of density will nearly be $.......\%$
- [NEET 2023]
A physical parameter a can be determined by measuring the parameters $b, c, d $ and $e $ using the relation $a =$ ${b^\alpha }{c^\beta }/{d^\gamma }{e^\delta }$. If the maximum errors in the measurement of $b, c, d$ and e are ${b_1}\%$, ${c_1}\%$, ${d_1}\%$ and ${e_1}\%$, then the maximum error in the value of a determined by the experiment is
A physical parameter a can be determined by measuring the parameters $b, c, d $ and $e $ using the relation $a =$ ${b^\alpha }{c^\beta }/{d^\gamma }{e^\delta }$. If the maximum errors in the measurement of $b, c, d$ and e are ${b_1}\%$, ${c_1}\%$, ${d_1}\%$ and ${e_1}\%$, then the maximum error in the value of a determined by the experiment is
Students $I$, $II$ and $III$ perform an experiment for measuring the acceleration due to gravity $(g)$ using a simple pendulum.
They use different lengths of the pendulum and /or record time for different number of oscillations. The observations are shown in the table.
Least count for length $=0.1 \mathrm{~cm}$
Least count for time $=0.1 \mathrm{~s}$
Student
Length of the pendulum $(cm)$
Number of oscillations $(n)$
Total time for $(n)$ oscillations $(s)$
Time period $(s)$
$I.$
$64.0$
$8$
$128.0$
$16.0$
$II.$
$64.0$
$4$
$64.0$
$16.0$
$III.$
$20.0$
$4$
$36.0$
$9.0$
If $\mathrm{E}_{\mathrm{I}}, \mathrm{E}_{\text {II }}$ and $\mathrm{E}_{\text {III }}$ are the percentage errors in g, i.e., $\left(\frac{\Delta \mathrm{g}}{\mathrm{g}} \times 100\right)$ for students $\mathrm{I}, \mathrm{II}$ and III, respectively,
Students $I$, $II$ and $III$ perform an experiment for measuring the acceleration due to gravity $(g)$ using a simple pendulum.
They use different lengths of the pendulum and /or record time for different number of oscillations. The observations are shown in the table.
Least count for length $=0.1 \mathrm{~cm}$
Least count for time $=0.1 \mathrm{~s}$
| Student | Length of the pendulum $(cm)$ | Number of oscillations $(n)$ | Total time for $(n)$ oscillations $(s)$ | Time period $(s)$ |
| $I.$ | $64.0$ | $8$ | $128.0$ | $16.0$ |
| $II.$ | $64.0$ | $4$ | $64.0$ | $16.0$ |
| $III.$ | $20.0$ | $4$ | $36.0$ | $9.0$ |
If $\mathrm{E}_{\mathrm{I}}, \mathrm{E}_{\text {II }}$ and $\mathrm{E}_{\text {III }}$ are the percentage errors in g, i.e., $\left(\frac{\Delta \mathrm{g}}{\mathrm{g}} \times 100\right)$ for students $\mathrm{I}, \mathrm{II}$ and III, respectively,
- [IIT 2008]