Ask & Answer Question

Physics NCERT Exemplar Solutions Class 12th Chapter Six Physics Class 12th Oscillations Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

An aluminium wire is stretched to make its length, 0.4% larger. The percentage change in resistance is

Option 1 -

0.4%

Option 2 -

0.2%

Option 3 -

0.8%

Option 4 -

0.6%

3 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

2 months ago

Correct Option - 3

Detailed Solution:

Volume = constant

$A l = C$

$A d l + l d A = 0$

$\frac{d l}{l} + \frac{d A}{A} = 0$

$\frac{d R}{R} \times 100 = (\frac{d l}{l} - \frac{d A}{A}) \times 100$

= 0.4 (0.4)

= 0.8 %

0 Upvotes 0 Downvotes

Similar Questions for you

In a vernier calipers, each cm on the main scale is divided into 20 equal parts. If tenth vernier scale division coincides with nineth main scale division. Then the value of vernier constant will be__________× 10^-2mm.

Payal Gupta

Let ‘x’ he the value of one division of main scale

$x = \frac{1}{20} c m = 0.05 c m$

Let y be value of one division on venire scale given

10 y = 9 x

$y = \frac{9 x}{10}$

Least count = $x - \frac{9 x}{10} = \frac{x}{10} = \frac{0.05}{10}$

= 0.005 cm

= 5 × 10^-2 mm

A particle executes S.H.M. with amplitude â€˜aâ€™ and time period â€˜Tâ€™. The displacement of the particle when its speed is half of maximum speed is $\frac{\sqrt[]{x} a}{2} .$ The value of x is

Payal Gupta

$v = ω \sqrt[]{A^{2} - s^{2}} \Rightarrow \frac{A ω}{2} = ω \sqrt[]{A^{2} - s^{2}} \Rightarrow s = \frac{\sqrt[]{3} A}{2} \Rightarrow x = 3$

Time period of a simple pendulum is T. The time taken to complete $\frac{5}{8}$ oscillations starting from mean position is $\frac{α}{β} T .$ The value of a is

Payal Gupta

Total oscillation = $\frac{1}{2} + \frac{1}{8} = \frac{5}{8}$

$\Rightarrow t i m e t a k e n = \frac{T}{2} + \frac{T}{12} = \frac{7 T}{12}$

= 7

When a particle executes Simple Hormonic Motion, the nature of graph of velocity as a function of displacement will be:

Vishal Baghel

x = A sin $ω t$

$\Rightarrow v = A ω c o s ω t$

$\Rightarrow v = \pm ω \sqrt[]{A^{2} - x^{2}}$

$\frac{v^{2}}{ω^{2}} + x^{2} = A^{2}$

Elliptical

Given below are two statements:

Statement I : A second’s pendulum has a time period of 1 second.

Statement II : It takes precisely one second to move between the two extreme positions.

In the light of the above statements, choose the correct answer from the options given below

Payal Gupta

Time period of second pendulum is 2 seconds.

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
687k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

An aluminium wire is stretched to make its length, 0.4% larger. The percentage change in resistance is

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question