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Physics-

General

Easy

Question

Mass M is displaced by a vertical distance $x subscript 0 end subscript$ from equilibrium position by applying a force F The spring constant are $k subscript 1 end subscript$ and $k subscript 2 end subscript open parentheses k subscript 1 end subscript greater than k subscript 2 end subscript close parentheses$ . Find the force F

$\frac{k_{1} k_{2}}{k_{1} + k_{2}} \cdot x_{0}$
$\frac{2 k_{1} k_{2}}{k_{1} + k_{2}} \cdot x_{0}$
$\frac{4 k_{1} k_{2}}{k_{1} + k_{2}} \cdot x_{0}$
$\frac{k_{1} k_{2}}{2 (k_{1} + k_{2})} \cdot x_{0}$

The correct answer is: $fraction numerator 4 k subscript 1 end subscript k subscript 2 end subscript over denominator k subscript 1 end subscript plus k subscript 2 end subscript end fraction times x subscript 0 end subscript$

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In the two systems as shown, what should be value of mass M so that extensions in two springs (of same spring constant k) is same (in kg ) ?

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Fig shows variation of force acting on a body, with time. Assuming the body to start from rest, the variation of its momentum with time is best represented by which plot?

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Find the mass M of the hanging block in figure. Which will prevent the smaller block from slipping over the triangular block. All the surfaces are frictionless and the strings and the pulleys are light.

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A block of mass $m$ is pulled by a string acts which passes over a smooth pulley P fitted with the wedge of mass M. A horizontal force F=mg acts on the string. If the string is massless and inextensible, assuming frictionless contacting surfaces and $theta equals c o s to the power of negative 1 end exponent invisible function application fraction numerator 4 over denominator 5 end fraction$ then

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