NEET Sample Paper NEET Sample Test Paper-68

  • question_answer
    A ball is dropped from a high rise platform at \[t\text{ }=0\] starting from rest. After 6 s another ball is thrown downwards from the same platform with a speed\[\nu \]. The two balls meet at\[\operatorname{t} =18 s\]. What is the value of\[\nu \]? \[\left( take\,\,g =10 m{{s}^{-}}^{2} \right)\]

    A) \[74\,\,m{{s}^{-}}^{2}\]                    

    B) \[55\,\,m{{s}^{-}}^{1}\]

    C) \[40\,\,m{{s}^{-}}^{1}\]                    

    D) \[60\,\,m{{s}^{-}}^{1}\]

    Correct Answer: A

    Solution :

    For first ball, \[\operatorname{u} = 0\] \[\therefore \,\,\,\,\,{{s}_{1}}=\frac{1}{2}\,gt_{1}^{2}=\frac{1}{2}\times g{{(18)}^{2}}\] For second ball, initial velocity = v \[\therefore \,\,\,\,\,{{s}_{2}}=\,\,\nu t_{2}^{{}}=\,\,\frac{1}{2}\,g{{t}^{2}}_{2}\] \[{{t}_{2}}\,=\,\,18-6\,\,=\,\,12\,s\] \[\Rightarrow \,{{s}_{2}}={{s}_{2}}=\nu \times 12+\frac{1}{2}\,g{{(12)}^{2}}\] Here, \[{{s}_{1}}={{s}_{2}}\] \[\frac{1}{2}g\left( 1{{8}^{2}} \right)=12\nu +\frac{1}{2}\,g{{\left( 12 \right)}^{2}}\,\,\,\Rightarrow \,\,\,\nu =74\,\,m{{s}^{-1}}\]


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