Tag: vernier calliper and screw gauge

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Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length
Unaware about the fact that analog ammeters and voltmeters can also have zero error, a student recorded following readings while determine resistance by Ohm's law 
If the ammeter has no zero error, the zero error is the voltmeter is 
 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      

  1. $-1 V$

  2. $-1.5V$

  3. $0.5V$

  4. $1V$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Ohm's law V = IR, the slope of the V-I graph gives the resistance. Plotting the points (1.0, 40), (3.0, 80), (5.0, 120), (7.0, 160), (9.0, 200) shows a linear relationship V = 0.05I - 1.0. When I = 0, V = -1.0V, indicating a zero error of -1V.

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

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

  1. $5\%$

  2. $4.5\%$

  3. $8\%$

  4. $7.75\%$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The relative error in Q is given by (dQ/Q) = (2/5)(dx/x) + (dy/y) + 3(dz/z) + (1/2)(dt/t). Substituting the given percentages: (2/5)*2.5% + 2% + 3*0.5% + (1/2)*1% = 1% + 2% + 1.5% + 0.5% = 5%.

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

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 $\left[ \alpha _ { \text { steel } } = 1.2 \times 10 ^ { - 5 \cdot 0 } C ^ { - 1 } \right]$

  1. $0.035 \%$

  2. $0.042 \%$

  3. $0.012 \%$

  4. $0.018 \%$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The change in length due to temperature is delta L = L * alpha * delta T. The percentage error is (delta L / L) * 100 = alpha * delta T * 100. Here, alpha = 1.2 * 10^-5 and delta T = 20 - (-15) = 35. So, 1.2 * 10^-5 * 35 * 100 = 0.042%.

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

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  

  1. 59 ${ \circ } _{ C }$

  2. 8.6 ${ \circ } _{ C }$

  3. 65.56 ${ \circ } _{ C }$

  4. 58 ${ \circ } _{ C }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

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

  1. $20$%

  2. $5$%

  3. $21$%

  4. $10$%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Kinetic energy K = p^2 / (2m). Since mass is constant, dK/K = 2 * dp/p. With dp/p = 10%, dK/K = 2 * 10% = 20%.

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

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

  1. $2\%$

  2. $3\%$

  3. $1\%$

  4. $1.5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

If the error in the measurement of radius of a sphere is 1%, then the error in the measurement of volume will be :

  1. 8%

  2. 5%

  3. 3%

  4. 1%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know that a sphere has volume $V=\dfrac{4}{3}\pi r^3$ meaning $V \alpha r^{3}$.
Thus, error in measurement of volume will be $ 3 \times$ 1% = 3%.

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

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

  1. $60$%

  2. $30$%

  3. $69$%

  4. $39$%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Kinetic energy K = (1/2)mv^2. Since mass is constant, dK/K = 2 * dv/v. Given dv/v = 30%, dK/K = 2 * 30% = 60%.

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

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 

  1. 2%

  2. 4%

  3. 7%

  4. 10%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given g = 4 * pi^2 * (L / T^2). The relative error is dg/g = dL/L + 2 * dT/T. Using magnitudes: 1% + 2 * 3% = 1% + 6% = 7%.