367268 If the time period \(t\) of the oscillation of a drop of liquid of density \(d,\) radius \(r,\) vibrating under surface tension \(s\) is given by the formula \(t=\sqrt{r^{2 b} s^{c} d^{a / 2}}\). It is observed that the time period is directly proportional \(\sqrt{\dfrac{d}{s}}\). The value of \(b\) should therefore be
367269 If \({d}\) is the depth to which a bullet of kinetic energy \({K}\) can penetrate into a human body of modulus of elasticity \({E}\), then using the method of dimension establish a relation between \({d, K}\) and \({E}\). If \({E=c d^{x} E^{y}}\), where \({c}\) is a dimensionless constant. Modulus of elasticity \({=\dfrac{\text { Force }}{\text { Area }} \times \dfrac{l}{\Delta l}}\) (where \({l}\) is length and \({\Delta l}\) is change in length). Find the value of \({x+y}\) is
367268 If the time period \(t\) of the oscillation of a drop of liquid of density \(d,\) radius \(r,\) vibrating under surface tension \(s\) is given by the formula \(t=\sqrt{r^{2 b} s^{c} d^{a / 2}}\). It is observed that the time period is directly proportional \(\sqrt{\dfrac{d}{s}}\). The value of \(b\) should therefore be
367269 If \({d}\) is the depth to which a bullet of kinetic energy \({K}\) can penetrate into a human body of modulus of elasticity \({E}\), then using the method of dimension establish a relation between \({d, K}\) and \({E}\). If \({E=c d^{x} E^{y}}\), where \({c}\) is a dimensionless constant. Modulus of elasticity \({=\dfrac{\text { Force }}{\text { Area }} \times \dfrac{l}{\Delta l}}\) (where \({l}\) is length and \({\Delta l}\) is change in length). Find the value of \({x+y}\) is
367268 If the time period \(t\) of the oscillation of a drop of liquid of density \(d,\) radius \(r,\) vibrating under surface tension \(s\) is given by the formula \(t=\sqrt{r^{2 b} s^{c} d^{a / 2}}\). It is observed that the time period is directly proportional \(\sqrt{\dfrac{d}{s}}\). The value of \(b\) should therefore be
367269 If \({d}\) is the depth to which a bullet of kinetic energy \({K}\) can penetrate into a human body of modulus of elasticity \({E}\), then using the method of dimension establish a relation between \({d, K}\) and \({E}\). If \({E=c d^{x} E^{y}}\), where \({c}\) is a dimensionless constant. Modulus of elasticity \({=\dfrac{\text { Force }}{\text { Area }} \times \dfrac{l}{\Delta l}}\) (where \({l}\) is length and \({\Delta l}\) is change in length). Find the value of \({x+y}\) is
367268 If the time period \(t\) of the oscillation of a drop of liquid of density \(d,\) radius \(r,\) vibrating under surface tension \(s\) is given by the formula \(t=\sqrt{r^{2 b} s^{c} d^{a / 2}}\). It is observed that the time period is directly proportional \(\sqrt{\dfrac{d}{s}}\). The value of \(b\) should therefore be
367269 If \({d}\) is the depth to which a bullet of kinetic energy \({K}\) can penetrate into a human body of modulus of elasticity \({E}\), then using the method of dimension establish a relation between \({d, K}\) and \({E}\). If \({E=c d^{x} E^{y}}\), where \({c}\) is a dimensionless constant. Modulus of elasticity \({=\dfrac{\text { Force }}{\text { Area }} \times \dfrac{l}{\Delta l}}\) (where \({l}\) is length and \({\Delta l}\) is change in length). Find the value of \({x+y}\) is