A screw gauge has some zero error but its value is unknown. We have two identical rods. When the first rod is inserted in the screw, the state of the instrument is shown by diagram (I). When both the rods are inserted together in series then the state is shown by the diagram (II). What is the zero error of the instrument? \(1~\text{msd}= 100~\text{csd}=1~\text{mm}\)

 

1. \(-0.16~\text{mm}\) 2. \(+0.16~\text{mm}\)
3. \(+0.14~\text{mm}\) 4. \(-0.14~\text{mm}\)

Subtopic:  Measurement & Measuring Devices |
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Find the zero correction in the given figure. 

 

1. \(0.4\) mm

2. \(0.5\) mm

3. \(-0.5\) mm

4. \(-0.4\) mm

Subtopic:  Measurement & Measuring Devices |
From NCERT
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Find the thickness of the wire. The least count is \(0.01~\text{mm}\). The main scale reads (in mm):

                       
1. \(7.62\)
2. \(7.63\)
3. \(7.64\)
4. \(7.65\)

Subtopic:  Measurement & Measuring Devices |
From NCERT
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The main scale reading is \(-1\) mm when there is no object between the jaws. In the vernier calipers, \(9\) main scale division matches with \(10\) vernier scale divisions. Assume the edge of the Vernier scale as the '\(0\)' of the vernier. The thickness of the object using the defected vernier calipers will be:

         
1. \(12.2~\text{mm}\)
2. \(1.22~\text{mm}\)
3. \(12.3~\text{mm}\)
4. \(12.4~\text{mm}\)

Subtopic:  Measurement & Measuring Devices |
 60%
From NCERT
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The main scale of a vernier callipers reads 10 mm in 10 divisions. 10 divisions of Vernier scale coincide with 9 divisions of the main scale. When the two jaws of the callipers touch each other, the fifth division of the vernier coincides with 9 main scale divisions and the zero of the vernier is to the right of zero of main scale. When a cylinder is tightly placed between the two jaws, the zero of vernier scale lies slightly behind 3.2 cm and the fourth vernier division coincides with a main scale division. The diameter of the cylinder is.

(1)  3.10 cm

(2)  3.8 cm

(3)  3.09 cm

(4)  -3.09 cm

Subtopic:  Measurement & Measuring Devices |
 51%
From NCERT
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A student measured the diameter of a small steel ball using a screw gauge of least count \(0.001\) cm. The main scale reading is \(5\) mm and zero of circular scale division coincides with \(25\) divisions above the reference level. If the screw gauge has a zero error of \(-0.004\) cm, the correct diameter of the ball is:
1. \(0.521\) cm 2. \(0.525\) cm
3. \(0.053\) cm 4. \(0.529\) cm
Subtopic:  Measurement & Measuring Devices |
 66%
From NCERT
NEET - 2018
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If energy \((E),\) velocity \((v)\) and time \((T)\) are chosen as the fundamental quantities, the dimensional formula of surface tension will be:
1. \([Ev^{-2}T^{-1}]\) 2. \([Ev^{-1}T^{-2}]\)
3. \([Ev^{-2}T^{-2}]\) 4. \([E^{-2}v^{-1}T^{-3}]\)

Subtopic:  Dimensions |
 70%
From NCERT
NEET - 2015
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In an experiment, four quantities \(a,\) \(b,\) \(c\) and \(d\) are measured with percentage error \(1\)%, \(2\)%, \(3\)% and \(4\)% respectively. Quantity \(P\) is calculated as follows: 
\(P=\dfrac{a^3b^2}{cd}\)
Percentage error in \(P\) is:
1. \(10\%\)
2. \(7\%\)
3. \(4\%\)
4. \(14\%\)
Subtopic:  Errors |
 90%
From NCERT
AIPMT - 2013
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The dimensions of μ0ε0-1/2 are:

1.  L-1 T

2.  LT-1

3.  L-1/2 T1/2

4.  L-1/2 T-1/2

Subtopic:  Dimensions |
 86%
From NCERT
AIPMT - 2011
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If the dimensions of a physical quantity are given by \([M^aL^bT^c],\) then the physical quantity will be:
1. pressure if \(a=1,\) \(b=-1,\) \(c=-2\)
2. velocity if \(a=1,\) \(b=0,\) \(c=-1\)
3. acceleration if \(a=1,\) \(b=1,\) \(c=-2\)
4. force if \(a=0,\) \(b=-1,\) \(c=-2\)
Subtopic:  Dimensions |
 87%
From NCERT
AIPMT - 2009
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