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3 months ago

Units and Measurement Class 11th

In an experiment to determine the Young's modulus of wire of a length exactly 1m, the extension in the length of the wire is measured as 0.4mm with an uncertainty of $\pm$ 0.02mm when a load of 1kg is applied. The diameter of the wire is measured as 0.4mm with an uncertainty of $\pm$ 0.02mm. The error in the measurement of Young's modulus $(Δ Y)$ is found to be x * 10¹⁰Nm^-2. The value of x is _________.

(take g = 10 ms^-2)

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Payal Gupta

Contributor-Level 10

L = 1 m

$Δ L = 0.4 * 1 0^{- 3} m$

$m = 1 k g$

d = 0.4 * 103 m

$\frac{F}{A} = y \frac{Δ L}{L}$

$y = \frac{F L}{A Δ L} = \frac{(m g) * 1}{(\frac{π d^{2}}{4}) * 0.4 * 1 0^{- 3}}$

$Δ y = 0.1 * 0.199 * 1 0^{12} = 1.99 * 1 0^{10}$

= 1.99

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3 months ago

Units and Measurement Class 11th

A screw gauge of pitch 0.5mm is used to measure the diameter of uniform wire of length 6.8cm, the main scale reading is 1.5mm and circular scale reading is 7. The calculated curved surface area wire to appropriate significant figures is:

[Screw gauge has 50 divisions on its circular scale]

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Payal Gupta

Contributor-Level 10

Least count = $\frac{0.5}{50} m m$ = 0.01 mm

Diameter, d = 1.5 + 7 * 0.01

= 1.57

$∴$ Surface Area = (2r) $l$

= 3.4 cm²

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3 months ago

Units and Measurement Class 11th

The quantities which have the same dimensions as those of solid angle are:

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alok kumar singh

Contributor-Level 10

Solid angle $d Ω = \frac{d A}{r^{2}}$ has dimensions $[M^{0} L^{0} T^{0}]$

Strain $= \frac{Δ l}{l}$ has dimensions $[M^{0} L^{0} T^{0}]$

Angle measured in radians is also dimensionless $[M^{0} L^{0} T^{0}]$

$θ = \frac{l}{r}$

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3 months ago

Units and Measurement Class 11th

Three students S₁, S₂ and S₂ perform an experiment for determining the acceleration due to gravity (g) using a simple pendulum. They use different lengths of pendulum and record time for different number of oscillations. The observations are as shown in the table.

Student No.	Length of Pendulum (cm)	No. of oscillations (n)	Total for n oscillations	Time period (s)
1	64.0	8	128.0	16.0
2	64.0	4	64.0	16.0
3	20.0	4	36.0	9.0

Least count of length = 0.1 cm

least count for time = 0.1 s)

If E₁, E₂ and E₃ are the percentage errors 'g' for students 1, 2 and 3 respectively, then the minimum percentage error is obtained by student no.___________.

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Vishal Baghel

Contributor-Level 10

$T = 2 π \sqrt[]{\frac{l}{g}}$

$g = 4 π^{2} \frac{l}{T^{2}}$

$\Rightarrow \frac{Δ g}{g} = \frac{Δ l}{l} + \frac{2 Δ T}{T}$

Student 1 will get least error because it has more number of oscillation

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3 months ago

Units and Measurement Class 11th

What are vernier calipers?

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Syed Aquib Ur Rahman

Contributor-Level 10

Vernier calipers are instruments for making accurate linear measurements. They are commonly used to measure the diameter of a sphere or the depth of a hole. Its precision comes from a sliding Vernier scale that runs along a fixed main scale.

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3 months ago

Units and Measurement Class 11th

What types of errors in measurement are there?

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Syed Aquib Ur Rahman

Contributor-Level 10

There are two types of errors in measurements. One is a systematic error, which comes from a fault in the measuring device. The other is a random error in measurement, which is caused by a sudden change in the environment.

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3 months ago

Units and Measurement Class 11th

If time (t), velocity $(υ)$ , and angular momentum $(l)$ are taken as the fundamental units. Then the dimension of mass (m) in terms of t, $υ, a n d l$ is:

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Vishal Baghel

Contributor-Level 10

Let Mass α $[t^{a} υ^{b} l^{c}] \Rightarrow [M^{1} L^{0} T^{0}] \equiv T^{a} * {[L T^{- 1}]}^{b} * {[M L^{2} T^{- 1}]}^{c}$

=> a – b – c = 0 c = 1, and b + 2c = 0 Þ b = -2c = -2

=> a = b + c = 1 – 2 = -1

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3 months ago

Units and Measurement Class 11th

The diameter of a spherical bob is measured using a vernier calipers. 9 divisions of the main scale, in the vernier caliper, are equal to 10 divisions of vernier scale. One main scale division is 1mm. The main scale reading is 10mm and 8^th division of vernier scale was found to coincide exactly with one of the main scale division. If the given vernier calipers has positive zero error of 0.04cm, then the radius of the bob is______ * 10^-2cm.

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Vishal Baghel

Contributor-Level 10

From question :

9MSD = 10VSD

LC = 1MSD – 1VSD = 1MSD - $\frac{9}{10} M S D = 0.1 M S D = 0.1 * 1 m m = 0.1 m m$

Reading = 10mm + 8 * 0.1 mm = 10.8 mm

Final reading = 10.8 mm – 0.04 mm = 10.4 mm = 104 * 10^-2 cm

So radius = 52 * 10^-2 cm

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3 months ago

Units and Measurement Class 11th

If velocity [V], time [T] and force [F] are chosen as the base quantities, the dimensions of the mass will be:

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Vishal Baghel

Contributor-Level 10

Using F = MA = $m \frac{V}{T}$

=> m = FTV^-1

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3 months ago

Units and Measurement Class 11th

Match List-I with List-II.

List-I List-II

(A) Torque (i) MLT^-1

...more

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Raj Pandey

Contributor-Level 9

(a) Torque ® ML²T^-2 ↑ (iii)

(b) Impulse ® MlT^-1 ↑ (i)

(d) Surface Tension ↑ $\frac{M L T^{- 2}}{L} = M T^{- 2}$ (ii)

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