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A massive rod of length 1m is placed on a smooth horizontal surface is pulled longitudinally by a force of 10N as shown in figure. The tension in rod varies as T=10√x.How will linear mass density vary with x?
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Asked by mailkunalmunjal | 09 Sep, 2020, 06:19: AM
answered-by-expert Expert Answer
Let a massive rod of 1 m length is subjected to a force 10 N as shown in figure. Let M be the mass of rod. 
 
Acceleration a of whole rod = Force / mass = ( 10 / M )
 
let us consider a small mass element dm at a distance x and having width dx.
 
Tension forces acting at forward and backward direction are shown in figure .
 
Forward direction tension force , TF = 10 begin mathsize 12px style square root of x plus d x end root end style 
Backward direction tension force, TB = 10 begin mathsize 12px style square root of x end style
By Newton's second law,  TF - TB = dm × a
 
Hence we have , begin mathsize 14px style 10 space square root of x space plus space d x end root space minus space 10 square root of x space equals space left parenthesis space lambda space d x space right parenthesis space left parenthesis 10 divided by M right parenthesis end style  ................(1)
where λ is linear mass density.
 
After simplification , eqn.(1) can be written as
 
begin mathsize 14px style x to the power of 1 divided by 2 end exponent space open parentheses 1 space plus space fraction numerator d x over denominator x end fraction close parentheses to the power of 1 divided by 2 end exponent space minus space x to the power of 1 divided by 2 end exponent space equals space lambda over M space d x end style
begin mathsize 14px style x to the power of 1 divided by 2 end exponent space open parentheses 1 space plus space 1 half fraction numerator d x over denominator x end fraction close parentheses space minus space x to the power of 1 divided by 2 end exponent space equals space 1 half space fraction numerator d x over denominator x to the power of 1 divided by 2 end exponent end fraction space equals space lambda over M space d x end style
Hence , λ = ( M/2 ) ( 1 / √x )
 
Hence linear mass density λ begin mathsize 14px style proportional to end style ( 1 / √x )
Answered by Thiyagarajan K | 15 Oct, 2020, 11:47: PM

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