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The breaking strength of the cable used to pull a body is 40 N. A body of mass 8 kg is resting on a table of coefficient of friction `mu=0.20`. The maximum acceleration which can be produced by the cable is ( take , `g=ms^(-2)` )A. `6 ms^(-2)`B. `3 ms^(-2)`C. `8 ms^(-2)`D. `8 ms^(-2)` |
Answer» Correct Answer - B (b) `a_("max")=(T_("max")-mu mg)/(m)=()40-0.2xx8xx(10)/(8)=3 ms^(-2)` |
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