141779
Two forces of magnitude \(F\) have a resultant of the same magnitude \(F\). The angle between the two forces is :
1 \(45^{\circ}\)
2 \(120^{\circ}\)
3 \(150^{\circ}\)
4 \(60^{\circ}\)
Explanation:
B As we know that force is vector quantity and it's resultant would be given as per the formula, \(\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\) Where, \(\theta=\) Angle between two forces. According to the question both the forces have magnitude \(\mathrm{F}\) and resultant \(\mathrm{F}\), So, \(\quad F=\sqrt{F^{2}+F^{2}+2 F \cdot F \cos \theta}\) \(\mathrm{F}^{2}=2 \mathrm{~F}^{2}+2 \mathrm{~F}^{2} \cos \theta\) \(-\mathrm{F}^{2}=2 \mathrm{~F}^{2} \cos \theta\) \(\cos \theta=-1 / 2\) \(\cos \theta=\cos 120^{\circ}\) \(\theta=120^{\circ}\) Hence, the force have same magnitude and same resultant then angle between them will be \(120^{\circ}\).
MP PET-2012
Motion in One Dimensions
141780
A Train accelerates from rest at a constant rate \(\alpha\) for distance \(x_{1}\) and time \(t_{1}\). After that it retards to rest at constant rate \(\beta\) for distance \(x_{2}\) and time \(t_{2}\). Which of the following relations is correct?
B So, \(\alpha=\frac{\mathrm{v}_{\max }}{\mathrm{t}_{1}}\) (i) \(\beta=\frac{v_{\max }}{t_{2}}\) \(\frac{\alpha}{\beta}=\frac{t_{2}}{t_{1}}\) Now, \(\mathrm{s}=\) area under \(\mathrm{v}-\mathrm{t}\) graph \(\mathrm{x}_{1}=\frac{1}{2} \times \mathrm{t}_{1} \times \mathrm{v}_{\mathrm{m}}\) \(\mathrm{x}_{2}=\frac{1}{2} \times \mathrm{t}_{2} \times \mathrm{v}_{\mathrm{m}}\) \(\frac{\mathrm{x}_{1}}{\mathrm{x}_{2}}=\frac{\mathrm{t}_{1}}{\mathrm{t}_{2}}=\frac{\beta}{\alpha}\)
MP PET-2012
Motion in One Dimensions
141781
An automobile travelling with a speed of 60 \(\mathrm{km} / \mathrm{h}\), can brake to stop within a distance of \(\mathbf{2 0 m}\). If the car is going twice as fast, i.e, 120 \(\mathrm{km} / \mathrm{h}\), the stopping distance will be
141782
A car moving with a speed of \(50 \mathrm{~km} / \mathrm{h}\), can be stopped by brakes after atleast \(6 \mathrm{~m}\). If the same car is moving at a speed of \(100 \mathrm{~km} / \mathrm{h}\), the minimum stopping distance is
1 \(12 \mathrm{~m}\)
2 \(18 \mathrm{~m}\)
3 \(24 \mathrm{~m}\)
4 \(6 \mathrm{~m}\)
Explanation:
C Given that, Speed of car, \(\left(\mathrm{u}_{1}\right)=50 \mathrm{~km} / \mathrm{h}\) and stopping distance \(\left(\mathrm{s}_{1}\right)\) \(=6 \mathrm{~m}\) Speed of car \(\left(\mathrm{u}_{2}\right)=100 \mathrm{~km} / \mathrm{h}\) According to Newton's third Law of motion, \(v^{2}=u^{2}-2 a s\) \(0=u^{2}-2 a s\) \(u^{2}=2 a s\) \(s \propto u^{2}\) [a is constant and negative as considering retardation] \(\therefore \frac{\mathrm{s}_{2}}{\mathrm{~s}_{1}}=\frac{\mathrm{u}_{2}^{2}}{\mathrm{u}_{1}^{2}}=\frac{100^{2}}{50^{2}}\) \(\mathrm{~s}_{2}=4 \mathrm{~s}_{1}=4 \times 6=24 \mathrm{~m}\)
141779
Two forces of magnitude \(F\) have a resultant of the same magnitude \(F\). The angle between the two forces is :
1 \(45^{\circ}\)
2 \(120^{\circ}\)
3 \(150^{\circ}\)
4 \(60^{\circ}\)
Explanation:
B As we know that force is vector quantity and it's resultant would be given as per the formula, \(\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\) Where, \(\theta=\) Angle between two forces. According to the question both the forces have magnitude \(\mathrm{F}\) and resultant \(\mathrm{F}\), So, \(\quad F=\sqrt{F^{2}+F^{2}+2 F \cdot F \cos \theta}\) \(\mathrm{F}^{2}=2 \mathrm{~F}^{2}+2 \mathrm{~F}^{2} \cos \theta\) \(-\mathrm{F}^{2}=2 \mathrm{~F}^{2} \cos \theta\) \(\cos \theta=-1 / 2\) \(\cos \theta=\cos 120^{\circ}\) \(\theta=120^{\circ}\) Hence, the force have same magnitude and same resultant then angle between them will be \(120^{\circ}\).
MP PET-2012
Motion in One Dimensions
141780
A Train accelerates from rest at a constant rate \(\alpha\) for distance \(x_{1}\) and time \(t_{1}\). After that it retards to rest at constant rate \(\beta\) for distance \(x_{2}\) and time \(t_{2}\). Which of the following relations is correct?
B So, \(\alpha=\frac{\mathrm{v}_{\max }}{\mathrm{t}_{1}}\) (i) \(\beta=\frac{v_{\max }}{t_{2}}\) \(\frac{\alpha}{\beta}=\frac{t_{2}}{t_{1}}\) Now, \(\mathrm{s}=\) area under \(\mathrm{v}-\mathrm{t}\) graph \(\mathrm{x}_{1}=\frac{1}{2} \times \mathrm{t}_{1} \times \mathrm{v}_{\mathrm{m}}\) \(\mathrm{x}_{2}=\frac{1}{2} \times \mathrm{t}_{2} \times \mathrm{v}_{\mathrm{m}}\) \(\frac{\mathrm{x}_{1}}{\mathrm{x}_{2}}=\frac{\mathrm{t}_{1}}{\mathrm{t}_{2}}=\frac{\beta}{\alpha}\)
MP PET-2012
Motion in One Dimensions
141781
An automobile travelling with a speed of 60 \(\mathrm{km} / \mathrm{h}\), can brake to stop within a distance of \(\mathbf{2 0 m}\). If the car is going twice as fast, i.e, 120 \(\mathrm{km} / \mathrm{h}\), the stopping distance will be
141782
A car moving with a speed of \(50 \mathrm{~km} / \mathrm{h}\), can be stopped by brakes after atleast \(6 \mathrm{~m}\). If the same car is moving at a speed of \(100 \mathrm{~km} / \mathrm{h}\), the minimum stopping distance is
1 \(12 \mathrm{~m}\)
2 \(18 \mathrm{~m}\)
3 \(24 \mathrm{~m}\)
4 \(6 \mathrm{~m}\)
Explanation:
C Given that, Speed of car, \(\left(\mathrm{u}_{1}\right)=50 \mathrm{~km} / \mathrm{h}\) and stopping distance \(\left(\mathrm{s}_{1}\right)\) \(=6 \mathrm{~m}\) Speed of car \(\left(\mathrm{u}_{2}\right)=100 \mathrm{~km} / \mathrm{h}\) According to Newton's third Law of motion, \(v^{2}=u^{2}-2 a s\) \(0=u^{2}-2 a s\) \(u^{2}=2 a s\) \(s \propto u^{2}\) [a is constant and negative as considering retardation] \(\therefore \frac{\mathrm{s}_{2}}{\mathrm{~s}_{1}}=\frac{\mathrm{u}_{2}^{2}}{\mathrm{u}_{1}^{2}}=\frac{100^{2}}{50^{2}}\) \(\mathrm{~s}_{2}=4 \mathrm{~s}_{1}=4 \times 6=24 \mathrm{~m}\)
141779
Two forces of magnitude \(F\) have a resultant of the same magnitude \(F\). The angle between the two forces is :
1 \(45^{\circ}\)
2 \(120^{\circ}\)
3 \(150^{\circ}\)
4 \(60^{\circ}\)
Explanation:
B As we know that force is vector quantity and it's resultant would be given as per the formula, \(\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\) Where, \(\theta=\) Angle between two forces. According to the question both the forces have magnitude \(\mathrm{F}\) and resultant \(\mathrm{F}\), So, \(\quad F=\sqrt{F^{2}+F^{2}+2 F \cdot F \cos \theta}\) \(\mathrm{F}^{2}=2 \mathrm{~F}^{2}+2 \mathrm{~F}^{2} \cos \theta\) \(-\mathrm{F}^{2}=2 \mathrm{~F}^{2} \cos \theta\) \(\cos \theta=-1 / 2\) \(\cos \theta=\cos 120^{\circ}\) \(\theta=120^{\circ}\) Hence, the force have same magnitude and same resultant then angle between them will be \(120^{\circ}\).
MP PET-2012
Motion in One Dimensions
141780
A Train accelerates from rest at a constant rate \(\alpha\) for distance \(x_{1}\) and time \(t_{1}\). After that it retards to rest at constant rate \(\beta\) for distance \(x_{2}\) and time \(t_{2}\). Which of the following relations is correct?
B So, \(\alpha=\frac{\mathrm{v}_{\max }}{\mathrm{t}_{1}}\) (i) \(\beta=\frac{v_{\max }}{t_{2}}\) \(\frac{\alpha}{\beta}=\frac{t_{2}}{t_{1}}\) Now, \(\mathrm{s}=\) area under \(\mathrm{v}-\mathrm{t}\) graph \(\mathrm{x}_{1}=\frac{1}{2} \times \mathrm{t}_{1} \times \mathrm{v}_{\mathrm{m}}\) \(\mathrm{x}_{2}=\frac{1}{2} \times \mathrm{t}_{2} \times \mathrm{v}_{\mathrm{m}}\) \(\frac{\mathrm{x}_{1}}{\mathrm{x}_{2}}=\frac{\mathrm{t}_{1}}{\mathrm{t}_{2}}=\frac{\beta}{\alpha}\)
MP PET-2012
Motion in One Dimensions
141781
An automobile travelling with a speed of 60 \(\mathrm{km} / \mathrm{h}\), can brake to stop within a distance of \(\mathbf{2 0 m}\). If the car is going twice as fast, i.e, 120 \(\mathrm{km} / \mathrm{h}\), the stopping distance will be
141782
A car moving with a speed of \(50 \mathrm{~km} / \mathrm{h}\), can be stopped by brakes after atleast \(6 \mathrm{~m}\). If the same car is moving at a speed of \(100 \mathrm{~km} / \mathrm{h}\), the minimum stopping distance is
1 \(12 \mathrm{~m}\)
2 \(18 \mathrm{~m}\)
3 \(24 \mathrm{~m}\)
4 \(6 \mathrm{~m}\)
Explanation:
C Given that, Speed of car, \(\left(\mathrm{u}_{1}\right)=50 \mathrm{~km} / \mathrm{h}\) and stopping distance \(\left(\mathrm{s}_{1}\right)\) \(=6 \mathrm{~m}\) Speed of car \(\left(\mathrm{u}_{2}\right)=100 \mathrm{~km} / \mathrm{h}\) According to Newton's third Law of motion, \(v^{2}=u^{2}-2 a s\) \(0=u^{2}-2 a s\) \(u^{2}=2 a s\) \(s \propto u^{2}\) [a is constant and negative as considering retardation] \(\therefore \frac{\mathrm{s}_{2}}{\mathrm{~s}_{1}}=\frac{\mathrm{u}_{2}^{2}}{\mathrm{u}_{1}^{2}}=\frac{100^{2}}{50^{2}}\) \(\mathrm{~s}_{2}=4 \mathrm{~s}_{1}=4 \times 6=24 \mathrm{~m}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Motion in One Dimensions
141779
Two forces of magnitude \(F\) have a resultant of the same magnitude \(F\). The angle between the two forces is :
1 \(45^{\circ}\)
2 \(120^{\circ}\)
3 \(150^{\circ}\)
4 \(60^{\circ}\)
Explanation:
B As we know that force is vector quantity and it's resultant would be given as per the formula, \(\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\) Where, \(\theta=\) Angle between two forces. According to the question both the forces have magnitude \(\mathrm{F}\) and resultant \(\mathrm{F}\), So, \(\quad F=\sqrt{F^{2}+F^{2}+2 F \cdot F \cos \theta}\) \(\mathrm{F}^{2}=2 \mathrm{~F}^{2}+2 \mathrm{~F}^{2} \cos \theta\) \(-\mathrm{F}^{2}=2 \mathrm{~F}^{2} \cos \theta\) \(\cos \theta=-1 / 2\) \(\cos \theta=\cos 120^{\circ}\) \(\theta=120^{\circ}\) Hence, the force have same magnitude and same resultant then angle between them will be \(120^{\circ}\).
MP PET-2012
Motion in One Dimensions
141780
A Train accelerates from rest at a constant rate \(\alpha\) for distance \(x_{1}\) and time \(t_{1}\). After that it retards to rest at constant rate \(\beta\) for distance \(x_{2}\) and time \(t_{2}\). Which of the following relations is correct?
B So, \(\alpha=\frac{\mathrm{v}_{\max }}{\mathrm{t}_{1}}\) (i) \(\beta=\frac{v_{\max }}{t_{2}}\) \(\frac{\alpha}{\beta}=\frac{t_{2}}{t_{1}}\) Now, \(\mathrm{s}=\) area under \(\mathrm{v}-\mathrm{t}\) graph \(\mathrm{x}_{1}=\frac{1}{2} \times \mathrm{t}_{1} \times \mathrm{v}_{\mathrm{m}}\) \(\mathrm{x}_{2}=\frac{1}{2} \times \mathrm{t}_{2} \times \mathrm{v}_{\mathrm{m}}\) \(\frac{\mathrm{x}_{1}}{\mathrm{x}_{2}}=\frac{\mathrm{t}_{1}}{\mathrm{t}_{2}}=\frac{\beta}{\alpha}\)
MP PET-2012
Motion in One Dimensions
141781
An automobile travelling with a speed of 60 \(\mathrm{km} / \mathrm{h}\), can brake to stop within a distance of \(\mathbf{2 0 m}\). If the car is going twice as fast, i.e, 120 \(\mathrm{km} / \mathrm{h}\), the stopping distance will be
141782
A car moving with a speed of \(50 \mathrm{~km} / \mathrm{h}\), can be stopped by brakes after atleast \(6 \mathrm{~m}\). If the same car is moving at a speed of \(100 \mathrm{~km} / \mathrm{h}\), the minimum stopping distance is
1 \(12 \mathrm{~m}\)
2 \(18 \mathrm{~m}\)
3 \(24 \mathrm{~m}\)
4 \(6 \mathrm{~m}\)
Explanation:
C Given that, Speed of car, \(\left(\mathrm{u}_{1}\right)=50 \mathrm{~km} / \mathrm{h}\) and stopping distance \(\left(\mathrm{s}_{1}\right)\) \(=6 \mathrm{~m}\) Speed of car \(\left(\mathrm{u}_{2}\right)=100 \mathrm{~km} / \mathrm{h}\) According to Newton's third Law of motion, \(v^{2}=u^{2}-2 a s\) \(0=u^{2}-2 a s\) \(u^{2}=2 a s\) \(s \propto u^{2}\) [a is constant and negative as considering retardation] \(\therefore \frac{\mathrm{s}_{2}}{\mathrm{~s}_{1}}=\frac{\mathrm{u}_{2}^{2}}{\mathrm{u}_{1}^{2}}=\frac{100^{2}}{50^{2}}\) \(\mathrm{~s}_{2}=4 \mathrm{~s}_{1}=4 \times 6=24 \mathrm{~m}\)
