Two bodies of mass 4 and 6 kg are attached to the ends of a string passing over a pulley. The 4 kg mass is attached to the table top by another string. Find the tension in the string T1. (g=9.8 m/s2).

T - 4g = 4a   --------equation 1
-T +6g = 6a  --------equation 2
Solving these two equations, we get
2g = 10a
a = 1.96m/s^2
m = 4+6=10kg
T1=10(1.96) = 19.6N
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t1+t=-4g ____. 1st eq
-t=6g. ------ 2nd eq
add eq 1 and eq 2
t= 2g
t=2(9.8)=19.6 m/s²
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Refer to the picture

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Refer to this picture it is correct

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Refer the above answers.
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By making free body diagram On mass of 4kg tension T1 will act as well as its weight i. e. 4g will act. Since the system is at rest so on 4kg object tension T=T1+4g------ eq 1 On mass of 6kg tension will equal to its weight i.e. T=6g ------2nd eq On solving will get 6g=T1+4g 6g-4g=T1 2g=T1 Given g=9.8m/s^2 Putting the value of g 2×9.8=T1 19.6N=T1
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the tension will be 19.6
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the answer will be 19.6 N
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Bhai 4kg pe 2 tension lag rahi hai string different  hai
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