5 months ago

Share

Answers : (2)

                    

Hi Nilamni,


What exactally you want to know in this?


 


Regards

2 years ago
                    As we know,
? = 1 2 L F µ ; w h e r e v i s f r e q u e n c y , F i s t h e t e n s i o n i n t h e s t r i n g a n d µ i s t h e m a s s d e n s i t y o f s t r i n g
According to Question,
288 = 1 2 × 80 F µ . . . . . . . . . . . . . . . ( 1 ) 312 = 1 2 L F µ . . . . . . . . . . . . . . . . . . . . ( 2 ) O n d i v i d i n g e q u a t i o n ( 1 ) a n d ( 2 ) w e g e t , 288 312 = 2 L 2 × 80 ? L = 73 . 85 c m
5 months ago

Post Your Answer

More Questions On Mechanics

Ask Experts

Have any Question? Ask Experts
Post Question
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!!
Click Here for details
A venturimeter equipped with a differential mercury manometer has an inlet diameter of 8cm and throat diameter is 4cm. Find rate of flow of water through the meter, if difference in mercury...
 
 
Let the velocity at inlet is v1 and at outlet is v2 by the equation of continuity A1v1 = A2v2 so v1/v2 = A2 / A1 = pi*r1^2 / pi*r2^2 = r1^2 / r2^2 = (4x10^-2)^2 / ( 2x10^-2)^2 = 16/4 = 4...
  img
Nirmal Singh. 3 months ago
 
Let the velocity at inlet is v1 and at outlet is v2 inlet diameterd1=8cm-->r1=4cm=0.04m outlet diameterd2=4cm-->r2=2cm=0.02m by the equation of continuity A1v1 = A2v2 so pr1²v1=pr2²v2...
 
chaitnyakishore 3 months ago
 
Sir you did the wrong calculation and according to your method, The right answer is, Let the velocity at inlet is v1 and at outlet is v2, Given that d1=8cm-->r1=4cm d2=4cm-->r2=2cm by the...
 
Pushkar Aditya 3 months ago
what are the units of torque ?
 
 
TORQUE= FORCE* RADIUS UNITS ARE N-m SHER MOHAMMAD B.TECH, IIT DELHI
  img
Sher Mohammad one month ago
 
In rotational equilibrium, the sum of the torques is equal to zero. In other words, there is no net torque on the object. Note that the SI units of torque is a Newton - metre
 
Harsha one month ago
A small block of mass m is kept at the left end of a larger block of mass M and length L which is on a horizontal surface such that the system can slide.The system is started towards right...
 
 
I think something is wrong in the question either the small block should be at the upper right corner of the bigger block or the system should be imparted velocity in the left direction.If...
 
Arpit Kumar 4 months ago
 
sq root. 4LM/gx(m+M)
 
RAJEEV CHOUDHARY 4 months ago
 
sq.rt. 4LM/gx(m+M)
 
RAJEEV CHOUDHARY 4 months ago
A transverse wave described by Y = (0.02 m) sin [(1.0 m-1) x + (30 s-1)t] Propagates on a stretched string having a linear mass density of 1.2 × 10-4 kg m-1. Find the tension in the string.
 
 
Sol. . m = mass per unit length = 1.2 * 10–4 kg/mt Y = (0.02m) sin [(1.0 m–1)x + (30 s–1)t] Here, k = 1 m–1 = 2π/λ ω = 30 s–1 = 2πf ∴ velocity of the wave in the stretched string v = λf =...
  img
Kevin Nash one month ago
In the determination of Young’s modulus Y=4MLg/pi*ld^2 by using Searle’s method, a wire of length L = 2mand diameter d = 0.5 mm is used. For a load M = 2.5 kg, an extension l = 0.25 mm in...
 
 
Hello Student, L.C. = 0.5/100= 0.005 mm ∆Y/y = ∆ll + 2∆(d)/d ∆l/l=0.005×10^{-3}/0.25×10{-3}=1/50 2∆d/d=2*0.005×10^{-3}/0.5×10{-3}=1/50 Thanks & Regards Arun Kumar...
  img
Arun Kumar 17 days ago
TWO POSITIVE POINT CHARGES PLACED AT A DISTANCE ‘b’ APART HAVE SUM Q.FOR WHAT VALUES OF THE CHARGES IS THE COULOMB FORCE BETWEEN THEM MAXIMUM?
 
 
METHOD 1: LET THE CHARGES ON THE TWO CHARGED PARTICLES BE Q 1 AND Q 2 ELECTROSTATIC FORCE BETWEEN THEM=K(Q 1 )(Q 2 )/b 2 USING A.M-G.M INEQUALITY FOR Q 1 Q 2 TO BE MAXIMUM , Q 1 =Q 2 i.e Q 1...
 
siddharth gupta 25 days ago
 
THANKS!
 
bharat makkar 25 days ago
View all Questions »