The pressure on a square plate is measured by measuring the force on the plate and length of the sides of the plate by using the formula P=\(\frac{F}{\iota ^{2}}\). If the maximum errors in the measurement of force and length are 4% and 2% respectively, then the maximum error in the measurement of pressure is

Solution

\(\frac{\Delta P}{P}\times 100=\frac{\Delta F}{F}\times 100+2\frac{\Delta \iota }{\iota }\times 100\)

=4% + 2 ×2% = 8%

The heat generated in a circuit is given by Q=I²Rt, where I is current, R is resistance and t is time. If the percentage errors in measuring I, R and t are 2%, 1% and1% respectively,then the maximum error in measuring heat will be

Solution

\(\frac{\Delta Q}{Q}\times 100=\frac{2\Delta I}{I}\times 100+\frac{\Delta R}{R}\times 100+\frac{\Delta t}{t}\times 100\)

= 2 × 2% + 1% + 1% + 1% = 6%

If x= a – b, then the maximum percentage error in the measurement of x will be

\(\left ( \frac{\Delta a}{a}+\frac{\Delta b}{b} \right )\times 100\)%
\(\left ( \frac{\Delta a}{a}-\frac{\Delta b}{b} \right )\times 100\)%
\(\left ( \frac{\Delta a}{a-b}-\frac{\Delta b}{a-b} \right )\times 100\)%
\(\left ( \frac{\Delta a}{a-b}+\frac{\Delta b}{a-b} \right )\times 100\)

Solution

Maximum absolute error is Δa+Δb. Therefore the

percentage error =\(\frac{absolute\, error}{actual\, value}\times 100\)

The pitch of the screw gauge is 0.5 mm. Its circular scale contains 50 divisions. The least count of the screw gauge is

0.001 mm
0.01 mm
0.02 mm
0.025 mm

Solution

Least count=\(\frac{0.5}{50}=0.01mm\)

Avernier calliper has 20 divisions on the vernier scale, which coincide with 19 on the main scale. The least count of the instrument is 0.1 mm.The main scale divisions are of

0.5 mm
1 mm
2mm
¹⁄₄mm

Solution

\(\frac{0.1}{10}=\left ( 1-\frac{19}{20} \right )MSD\Rightarrow \frac{1}{100}=\frac{1}{20}\times 1MSD\)

⇒1 MSD =¹⁄₅×10=2

In a vernier callipers, ten smallest divisions of the vernier scale are equal to nine smallest division on the main scale. If the smallest division on the main scale is half millimeter,then the vernier constant is

0.5 mm
0.1mm
0.05 mm
0.005 mm

Solution

10 VD = 9MD, 1VD =\(\frac{9}{10}MD\)

Vernier constant =1 MD – 1 VD

\(\left ( 1-\frac{9}{10} \right )MD=\frac{1}{10}MD=\frac{1}{10}\times \frac{1}{2}=0.05 mm\)

If E, m, J and G represent energy, mass, angular momentum and gravitational constant respectively,then the dimensional formula of EJ²/ m⁵ G² is

angle
length
mass
time

Solution

\(\frac{[ML^{2}T^{-2}][ML^{2}T^{-1}]^{2}}{[M^{5}][M^{-1}L^{3} T^{-2}]^{2}}=[M^{0}L^{0}T^{0}]=angel\)

The thrust developed by a rocket-motor is given by F=mv+A(P₁P₂) where m is the mass of the gas ejected per unit time, v is velocity of the gas, A is area of cross-section of the nozzle, P₁ and P₂ are the pressures of the exhaust gas and surrounding atmosphere. The formula is dimensionally

correct
wrong
sometimes wrong, sometimes correct
Data is not adequate

Solution

Use principle of homogeneity.

The dimensional formula of velocity gradient is

[M⁰L⁰T^-1]
[MLT¹]
[ML²T^-1]
[M⁰LT^-2]

Solution

Velocity gradient is velocity per unit distance.

Turpentine oil is flowing through a capillary tube of length ι and radius r. The pressure difference between the two ends of the tube is p. The viscosity of oil is given by :\(\eta =\frac{p(r^{2}-x^{2})}{4v\iota }\). Here v is velocity of oil at a distance x from the axis of the tube. From this relation, the dimensional formula of η is

[ML^-1T^-1]
[MLT^-1]
[ML²T²]
[M⁰L⁰T⁰]

Solution

η is the coefficient of viscosity.

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