372193
Keeping the banking angle same, to increase the maximum speed with which a vehicle can travel on a curved road by 10 percent the radius of curvature of the road has to be changed from \(20 \mathrm{~m}\) to
1 \(6 \mathrm{~m}\)
2 \(18 \mathrm{~m}\)
3 \(24.2 \mathrm{~m}\)
4 \(30.5 \mathrm{~m}\)
Explanation:
C For speed v, we have \(\tan \theta=\frac{\mathrm{v}^{2}}{\mathrm{rg}}=\frac{\mathrm{v}^{2}}{20 \mathrm{~g}}\) If \(\quad \mathrm{v}_{2}=\mathrm{v}+0.1 \mathrm{v}=1.1 \mathrm{v}\) then, \(\tan \theta=\frac{\mathrm{v}_{2}^{2}}{\mathrm{r}_{2} \mathrm{~g}}=\frac{(1.1 \mathrm{v})^{2}}{\mathrm{r}_{2} \mathrm{~g}}\) Since the banking angle is constant, Equating \(\mathrm{eq}^{\mathrm{n}}\) (i) and (ii) \(\frac{1}{20}=\frac{1.21}{r_{2}}\) \(r_{2}=20 \times 1.21=24.2 \mathrm{~m}\)
EAMCET-1991
LAWS OF MOTION (ADDITIONAL)
372194
The banking angle for a curved road of radius \(490 \mathrm{~m}\) for a vehicle moving at \(35 \mathrm{~ms}^{-1}\) is
1 \(\tan ^{-1}(0.25)\)
2 \(\tan ^{-1}(0.55)\)
3 \(\tan ^{-1}(0.45)\)
4 \(\tan ^{-1}(0.75)\)
Explanation:
A Here, \(v=35 \mathrm{~ms}^{-1}, \quad \mathrm{r}=490 \mathrm{~m}\) \(\therefore \tan \theta =\frac{\mathrm{v}^{2}}{\operatorname{rg}}=\frac{(35)^{2}}{490 \times 9.8}=0.25\) \(\text { or } \theta =\tan ^{-1}(0.25)\)
EAMCET-1992
LAWS OF MOTION (ADDITIONAL)
372195
The maximum speed with which a car can be driven round a curve path of a radius \(18 \mathrm{~m}\) without skidding (when \(\mathrm{g}=10 \mathrm{~ms}^{-2}\) and the coefficient of friction between rubber tyres and the roadways is 0.2 ) is
1 \(36.0 \mathrm{kmh}^{-1}\)
2 \(18.0 \mathrm{kmh}^{-1}\)
3 \(21.6 \mathrm{kmh}^{-1}\)
4 \(14.4 \mathrm{kmh}^{-1}\)
Explanation:
C Centripetal force \(=\) Frictional force \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mu \mathrm{mg}\) \(\mathrm{v}=\sqrt{\mu \mathrm{rg}}\) \(\mathrm{v}=\sqrt{0.2 \times 18 \times 10}\) \(\mathrm{v}=6 \mathrm{~m} / \mathrm{s}\) \(\mathrm{v}=21.6 \mathrm{kmph}\)
EAMCET-1998
LAWS OF MOTION (ADDITIONAL)
372196
Two particles \(A\) and \(B\) are connected by a rigid rod \(A B\). The rod slides along perpendicular rails as shown here. The velocity of \(A\) to the right is \(10 \mathrm{~m} / \mathrm{s}\). What is the velocity of \(B\) when angle \(\alpha=60^{\circ}\) ?
1 \(9.8 \mathrm{~m} / \mathrm{s}\)
2 \(10 \mathrm{~m} / \mathrm{s}\)
3 \(5.8 \mathrm{~m} / \mathrm{s}\)
4 \(17.3 \mathrm{~m} / \mathrm{s}\)
Explanation:
D Given, \(\mathrm{v}_{\mathrm{x}}=10 \mathrm{~m} / \mathrm{sec}\) Angle \((\alpha)=60^{\circ}\) Let the velocity along \(\mathrm{x}\) and \(\mathrm{y}\) axes be \(\mathrm{v}_{\mathrm{x}}\) and \(\mathrm{v}_{\mathrm{y}}\) respectively, \(\therefore \quad \mathrm{v}_{\mathrm{x}}=\frac{\mathrm{dx}}{\mathrm{dt}} \text { and } \mathrm{v}_{\mathrm{y}}=\frac{\mathrm{dy}}{\mathrm{dt}}\) From figure, \(\tan \alpha=y / x\) \(\Rightarrow \quad y=x \tan \alpha\) Differentiating equation (i) with respect to \(t\), we get- \(\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{dx}}{\mathrm{dt}} \tan \alpha\) \(\mathrm{v}_{\mathrm{y}}=\mathrm{v}_{\mathrm{x}} \tan 60^{\circ}\) \(\mathrm{v}_{\mathrm{y}}=10 \times \sqrt{3}\) \(\mathrm{v}_{\mathrm{y}}=10 \times 1.731\) \(\mathrm{v}_{\mathrm{y}}=17.31 \mathrm{~m} / \mathrm{sec}\)
372193
Keeping the banking angle same, to increase the maximum speed with which a vehicle can travel on a curved road by 10 percent the radius of curvature of the road has to be changed from \(20 \mathrm{~m}\) to
1 \(6 \mathrm{~m}\)
2 \(18 \mathrm{~m}\)
3 \(24.2 \mathrm{~m}\)
4 \(30.5 \mathrm{~m}\)
Explanation:
C For speed v, we have \(\tan \theta=\frac{\mathrm{v}^{2}}{\mathrm{rg}}=\frac{\mathrm{v}^{2}}{20 \mathrm{~g}}\) If \(\quad \mathrm{v}_{2}=\mathrm{v}+0.1 \mathrm{v}=1.1 \mathrm{v}\) then, \(\tan \theta=\frac{\mathrm{v}_{2}^{2}}{\mathrm{r}_{2} \mathrm{~g}}=\frac{(1.1 \mathrm{v})^{2}}{\mathrm{r}_{2} \mathrm{~g}}\) Since the banking angle is constant, Equating \(\mathrm{eq}^{\mathrm{n}}\) (i) and (ii) \(\frac{1}{20}=\frac{1.21}{r_{2}}\) \(r_{2}=20 \times 1.21=24.2 \mathrm{~m}\)
EAMCET-1991
LAWS OF MOTION (ADDITIONAL)
372194
The banking angle for a curved road of radius \(490 \mathrm{~m}\) for a vehicle moving at \(35 \mathrm{~ms}^{-1}\) is
1 \(\tan ^{-1}(0.25)\)
2 \(\tan ^{-1}(0.55)\)
3 \(\tan ^{-1}(0.45)\)
4 \(\tan ^{-1}(0.75)\)
Explanation:
A Here, \(v=35 \mathrm{~ms}^{-1}, \quad \mathrm{r}=490 \mathrm{~m}\) \(\therefore \tan \theta =\frac{\mathrm{v}^{2}}{\operatorname{rg}}=\frac{(35)^{2}}{490 \times 9.8}=0.25\) \(\text { or } \theta =\tan ^{-1}(0.25)\)
EAMCET-1992
LAWS OF MOTION (ADDITIONAL)
372195
The maximum speed with which a car can be driven round a curve path of a radius \(18 \mathrm{~m}\) without skidding (when \(\mathrm{g}=10 \mathrm{~ms}^{-2}\) and the coefficient of friction between rubber tyres and the roadways is 0.2 ) is
1 \(36.0 \mathrm{kmh}^{-1}\)
2 \(18.0 \mathrm{kmh}^{-1}\)
3 \(21.6 \mathrm{kmh}^{-1}\)
4 \(14.4 \mathrm{kmh}^{-1}\)
Explanation:
C Centripetal force \(=\) Frictional force \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mu \mathrm{mg}\) \(\mathrm{v}=\sqrt{\mu \mathrm{rg}}\) \(\mathrm{v}=\sqrt{0.2 \times 18 \times 10}\) \(\mathrm{v}=6 \mathrm{~m} / \mathrm{s}\) \(\mathrm{v}=21.6 \mathrm{kmph}\)
EAMCET-1998
LAWS OF MOTION (ADDITIONAL)
372196
Two particles \(A\) and \(B\) are connected by a rigid rod \(A B\). The rod slides along perpendicular rails as shown here. The velocity of \(A\) to the right is \(10 \mathrm{~m} / \mathrm{s}\). What is the velocity of \(B\) when angle \(\alpha=60^{\circ}\) ?
1 \(9.8 \mathrm{~m} / \mathrm{s}\)
2 \(10 \mathrm{~m} / \mathrm{s}\)
3 \(5.8 \mathrm{~m} / \mathrm{s}\)
4 \(17.3 \mathrm{~m} / \mathrm{s}\)
Explanation:
D Given, \(\mathrm{v}_{\mathrm{x}}=10 \mathrm{~m} / \mathrm{sec}\) Angle \((\alpha)=60^{\circ}\) Let the velocity along \(\mathrm{x}\) and \(\mathrm{y}\) axes be \(\mathrm{v}_{\mathrm{x}}\) and \(\mathrm{v}_{\mathrm{y}}\) respectively, \(\therefore \quad \mathrm{v}_{\mathrm{x}}=\frac{\mathrm{dx}}{\mathrm{dt}} \text { and } \mathrm{v}_{\mathrm{y}}=\frac{\mathrm{dy}}{\mathrm{dt}}\) From figure, \(\tan \alpha=y / x\) \(\Rightarrow \quad y=x \tan \alpha\) Differentiating equation (i) with respect to \(t\), we get- \(\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{dx}}{\mathrm{dt}} \tan \alpha\) \(\mathrm{v}_{\mathrm{y}}=\mathrm{v}_{\mathrm{x}} \tan 60^{\circ}\) \(\mathrm{v}_{\mathrm{y}}=10 \times \sqrt{3}\) \(\mathrm{v}_{\mathrm{y}}=10 \times 1.731\) \(\mathrm{v}_{\mathrm{y}}=17.31 \mathrm{~m} / \mathrm{sec}\)
372193
Keeping the banking angle same, to increase the maximum speed with which a vehicle can travel on a curved road by 10 percent the radius of curvature of the road has to be changed from \(20 \mathrm{~m}\) to
1 \(6 \mathrm{~m}\)
2 \(18 \mathrm{~m}\)
3 \(24.2 \mathrm{~m}\)
4 \(30.5 \mathrm{~m}\)
Explanation:
C For speed v, we have \(\tan \theta=\frac{\mathrm{v}^{2}}{\mathrm{rg}}=\frac{\mathrm{v}^{2}}{20 \mathrm{~g}}\) If \(\quad \mathrm{v}_{2}=\mathrm{v}+0.1 \mathrm{v}=1.1 \mathrm{v}\) then, \(\tan \theta=\frac{\mathrm{v}_{2}^{2}}{\mathrm{r}_{2} \mathrm{~g}}=\frac{(1.1 \mathrm{v})^{2}}{\mathrm{r}_{2} \mathrm{~g}}\) Since the banking angle is constant, Equating \(\mathrm{eq}^{\mathrm{n}}\) (i) and (ii) \(\frac{1}{20}=\frac{1.21}{r_{2}}\) \(r_{2}=20 \times 1.21=24.2 \mathrm{~m}\)
EAMCET-1991
LAWS OF MOTION (ADDITIONAL)
372194
The banking angle for a curved road of radius \(490 \mathrm{~m}\) for a vehicle moving at \(35 \mathrm{~ms}^{-1}\) is
1 \(\tan ^{-1}(0.25)\)
2 \(\tan ^{-1}(0.55)\)
3 \(\tan ^{-1}(0.45)\)
4 \(\tan ^{-1}(0.75)\)
Explanation:
A Here, \(v=35 \mathrm{~ms}^{-1}, \quad \mathrm{r}=490 \mathrm{~m}\) \(\therefore \tan \theta =\frac{\mathrm{v}^{2}}{\operatorname{rg}}=\frac{(35)^{2}}{490 \times 9.8}=0.25\) \(\text { or } \theta =\tan ^{-1}(0.25)\)
EAMCET-1992
LAWS OF MOTION (ADDITIONAL)
372195
The maximum speed with which a car can be driven round a curve path of a radius \(18 \mathrm{~m}\) without skidding (when \(\mathrm{g}=10 \mathrm{~ms}^{-2}\) and the coefficient of friction between rubber tyres and the roadways is 0.2 ) is
1 \(36.0 \mathrm{kmh}^{-1}\)
2 \(18.0 \mathrm{kmh}^{-1}\)
3 \(21.6 \mathrm{kmh}^{-1}\)
4 \(14.4 \mathrm{kmh}^{-1}\)
Explanation:
C Centripetal force \(=\) Frictional force \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mu \mathrm{mg}\) \(\mathrm{v}=\sqrt{\mu \mathrm{rg}}\) \(\mathrm{v}=\sqrt{0.2 \times 18 \times 10}\) \(\mathrm{v}=6 \mathrm{~m} / \mathrm{s}\) \(\mathrm{v}=21.6 \mathrm{kmph}\)
EAMCET-1998
LAWS OF MOTION (ADDITIONAL)
372196
Two particles \(A\) and \(B\) are connected by a rigid rod \(A B\). The rod slides along perpendicular rails as shown here. The velocity of \(A\) to the right is \(10 \mathrm{~m} / \mathrm{s}\). What is the velocity of \(B\) when angle \(\alpha=60^{\circ}\) ?
1 \(9.8 \mathrm{~m} / \mathrm{s}\)
2 \(10 \mathrm{~m} / \mathrm{s}\)
3 \(5.8 \mathrm{~m} / \mathrm{s}\)
4 \(17.3 \mathrm{~m} / \mathrm{s}\)
Explanation:
D Given, \(\mathrm{v}_{\mathrm{x}}=10 \mathrm{~m} / \mathrm{sec}\) Angle \((\alpha)=60^{\circ}\) Let the velocity along \(\mathrm{x}\) and \(\mathrm{y}\) axes be \(\mathrm{v}_{\mathrm{x}}\) and \(\mathrm{v}_{\mathrm{y}}\) respectively, \(\therefore \quad \mathrm{v}_{\mathrm{x}}=\frac{\mathrm{dx}}{\mathrm{dt}} \text { and } \mathrm{v}_{\mathrm{y}}=\frac{\mathrm{dy}}{\mathrm{dt}}\) From figure, \(\tan \alpha=y / x\) \(\Rightarrow \quad y=x \tan \alpha\) Differentiating equation (i) with respect to \(t\), we get- \(\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{dx}}{\mathrm{dt}} \tan \alpha\) \(\mathrm{v}_{\mathrm{y}}=\mathrm{v}_{\mathrm{x}} \tan 60^{\circ}\) \(\mathrm{v}_{\mathrm{y}}=10 \times \sqrt{3}\) \(\mathrm{v}_{\mathrm{y}}=10 \times 1.731\) \(\mathrm{v}_{\mathrm{y}}=17.31 \mathrm{~m} / \mathrm{sec}\)
372193
Keeping the banking angle same, to increase the maximum speed with which a vehicle can travel on a curved road by 10 percent the radius of curvature of the road has to be changed from \(20 \mathrm{~m}\) to
1 \(6 \mathrm{~m}\)
2 \(18 \mathrm{~m}\)
3 \(24.2 \mathrm{~m}\)
4 \(30.5 \mathrm{~m}\)
Explanation:
C For speed v, we have \(\tan \theta=\frac{\mathrm{v}^{2}}{\mathrm{rg}}=\frac{\mathrm{v}^{2}}{20 \mathrm{~g}}\) If \(\quad \mathrm{v}_{2}=\mathrm{v}+0.1 \mathrm{v}=1.1 \mathrm{v}\) then, \(\tan \theta=\frac{\mathrm{v}_{2}^{2}}{\mathrm{r}_{2} \mathrm{~g}}=\frac{(1.1 \mathrm{v})^{2}}{\mathrm{r}_{2} \mathrm{~g}}\) Since the banking angle is constant, Equating \(\mathrm{eq}^{\mathrm{n}}\) (i) and (ii) \(\frac{1}{20}=\frac{1.21}{r_{2}}\) \(r_{2}=20 \times 1.21=24.2 \mathrm{~m}\)
EAMCET-1991
LAWS OF MOTION (ADDITIONAL)
372194
The banking angle for a curved road of radius \(490 \mathrm{~m}\) for a vehicle moving at \(35 \mathrm{~ms}^{-1}\) is
1 \(\tan ^{-1}(0.25)\)
2 \(\tan ^{-1}(0.55)\)
3 \(\tan ^{-1}(0.45)\)
4 \(\tan ^{-1}(0.75)\)
Explanation:
A Here, \(v=35 \mathrm{~ms}^{-1}, \quad \mathrm{r}=490 \mathrm{~m}\) \(\therefore \tan \theta =\frac{\mathrm{v}^{2}}{\operatorname{rg}}=\frac{(35)^{2}}{490 \times 9.8}=0.25\) \(\text { or } \theta =\tan ^{-1}(0.25)\)
EAMCET-1992
LAWS OF MOTION (ADDITIONAL)
372195
The maximum speed with which a car can be driven round a curve path of a radius \(18 \mathrm{~m}\) without skidding (when \(\mathrm{g}=10 \mathrm{~ms}^{-2}\) and the coefficient of friction between rubber tyres and the roadways is 0.2 ) is
1 \(36.0 \mathrm{kmh}^{-1}\)
2 \(18.0 \mathrm{kmh}^{-1}\)
3 \(21.6 \mathrm{kmh}^{-1}\)
4 \(14.4 \mathrm{kmh}^{-1}\)
Explanation:
C Centripetal force \(=\) Frictional force \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mu \mathrm{mg}\) \(\mathrm{v}=\sqrt{\mu \mathrm{rg}}\) \(\mathrm{v}=\sqrt{0.2 \times 18 \times 10}\) \(\mathrm{v}=6 \mathrm{~m} / \mathrm{s}\) \(\mathrm{v}=21.6 \mathrm{kmph}\)
EAMCET-1998
LAWS OF MOTION (ADDITIONAL)
372196
Two particles \(A\) and \(B\) are connected by a rigid rod \(A B\). The rod slides along perpendicular rails as shown here. The velocity of \(A\) to the right is \(10 \mathrm{~m} / \mathrm{s}\). What is the velocity of \(B\) when angle \(\alpha=60^{\circ}\) ?
1 \(9.8 \mathrm{~m} / \mathrm{s}\)
2 \(10 \mathrm{~m} / \mathrm{s}\)
3 \(5.8 \mathrm{~m} / \mathrm{s}\)
4 \(17.3 \mathrm{~m} / \mathrm{s}\)
Explanation:
D Given, \(\mathrm{v}_{\mathrm{x}}=10 \mathrm{~m} / \mathrm{sec}\) Angle \((\alpha)=60^{\circ}\) Let the velocity along \(\mathrm{x}\) and \(\mathrm{y}\) axes be \(\mathrm{v}_{\mathrm{x}}\) and \(\mathrm{v}_{\mathrm{y}}\) respectively, \(\therefore \quad \mathrm{v}_{\mathrm{x}}=\frac{\mathrm{dx}}{\mathrm{dt}} \text { and } \mathrm{v}_{\mathrm{y}}=\frac{\mathrm{dy}}{\mathrm{dt}}\) From figure, \(\tan \alpha=y / x\) \(\Rightarrow \quad y=x \tan \alpha\) Differentiating equation (i) with respect to \(t\), we get- \(\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{dx}}{\mathrm{dt}} \tan \alpha\) \(\mathrm{v}_{\mathrm{y}}=\mathrm{v}_{\mathrm{x}} \tan 60^{\circ}\) \(\mathrm{v}_{\mathrm{y}}=10 \times \sqrt{3}\) \(\mathrm{v}_{\mathrm{y}}=10 \times 1.731\) \(\mathrm{v}_{\mathrm{y}}=17.31 \mathrm{~m} / \mathrm{sec}\)
