I think ques may value Ys ki likhi 20... Jb kay hmy kha ha jo force provide kr ri shear strain Agr book kay page 114 pr dky to shear kay liay hmy jo value ha wo 8 .0 HA × 10 ^ 10 AGR SLOVE KRIN THEN 80 × 10 ^ 9 A JATA IS TRHA ANS SLOVE KR KAY SARA AT THE END 5.2 AY GA IN SHORT BOOK KAY NUMERICAL MAY VALUE WRONG HA YS: 20× 10 ^ 10 ❌ Shear .. : 80 × 10 ^ 9💯
I think probably the reason for this is that we don't have the values to use formulas which are used in previous numericals that is why we used another formula
1st problem main Y ki value 8×10^10 looo....bcz length main to change NAHI aa Raha yeh ....agr ap ke pas nbf ki book hai to pg 114 pr Jo table hai uss ko dekh lo
Actually according to the given information in numerical the mass is 13 kg as the Ys in given data is 20×10*10 See table 7.1 the shear Ys value is 8×10*10 if we put this then the mass will be 5.2kg I would say that Ys value is given wrong in numerical it should be 8×10*10 as we r talking about shear strain
I think ques may value Ys ki likhi 20...
Jb kay hmy kha ha jo force provide kr ri shear strain
Agr book kay page 114 pr dky to shear kay liay hmy jo value ha wo 8 .0 HA × 10 ^ 10 AGR SLOVE KRIN
THEN 80 × 10 ^ 9 A JATA IS TRHA ANS SLOVE KR KAY SARA AT THE END 5.2 AY GA
IN SHORT BOOK KAY NUMERICAL MAY VALUE WRONG HA
YS: 20× 10 ^ 10 ❌
Shear .. : 80 × 10 ^ 9💯
at last why using that formula while in second last we have used any other formula for strain energy
Thanks alot ❤
Sir last one me apny jo formula use kiya ku kya he second last me tu dosra use kiya tha strain energy ka
Yes sir how the formula is changed?
I think probably the reason for this is that we don't have the values to use formulas which are used in previous numericals that is why we used another formula
book answers doesn't match
1st problem main Y ki value 8×10^10 looo....bcz length main to change NAHI aa Raha yeh ....agr ap ke pas nbf ki book hai to pg 114 pr Jo table hai uss ko dekh lo
First numerical answer is 5.2kg in book sir please tell which one is right
Book answer is wrong
it is right
On the keybook they use the formula
Delta x = 1f/SA *L
Actually according to the given information in numerical the mass is 13 kg as the Ys in given data is 20×10*10
See table 7.1 the shear Ys value is 8×10*10 if we put this then the mass will be 5.2kg
I would say that Ys value is given wrong in numerical it should be 8×10*10 as we r talking about shear strain
Why the answer's are not according to the book???
Kuka mostly answer book per ghlat ha
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