wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Three blocks of masses 2 kg, 4 kg and 6 kg are connected by strings and placed on a fricitionless incline of 53 as shown. A force of 120 N is applied upward along the incline on the 6 kg block so that system accelerates. If the strings are ideal, the ratio tension in the strings T1T2 will be (g=10 ms2)



A
1:1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1:2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
1:3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
1:4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 1:3
It is given that the masses are,

m1=2 kg, m2=4 kg, m3=6 kg

hence we know that the acceleration of the system will be

asys=applied forceopposing forcetotal mass

Opposing force will be component of weight along the incline. So, opposing force=12g×sin53o down the incline.

Hence acceleration will be,
asys=12012×10×4512=2412=2 ms2

By the free body diagram of the blocks, we get



Now writing force equation for the block of mass m1=2 kg

T1m1gsin53=m1asys, T1=2×2+2×10×45=20 N

Now writing force equation for the block m3=6 kg

FT2m3gsin53o=6asys120T26×10×45=6×2

We get T2=60 N and T1=20 N

T1T2=13

Hence option C is the correct answer.

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
What's with All the Tension in the Room?
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon