Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

Two springs of spring constants k1 and k2 are joined in series. The effective spring constant of the-combination is given by:

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya

a

k1k2

b

(k1+k2)2

c

k1+k2

d

k1k2(k1+k2)

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

Let us consider two springs of spring constants k1 and k2 joined in series as shown in the figure.Under a force F, they will stretch by y1 and y2.So y = y1+y2or Fk  = F1k1+F2k2but as springs are massless, so force on them must be the same, i,e., F1 = F2 = FSo, 1k = 1k1+1k2     or  k = k1k2k1+k2Aliter: In series combination,1kS = 1k1+1k2 = k2+k1k1k2     ⇒kS = k1k2k1+k2
Watch 3-min video & get full concept clarity

courses

No courses found

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon