{"id":591,"date":"2019-08-14T16:44:48","date_gmt":"2019-08-14T11:14:48","guid":{"rendered":"https:\/\/www.physicsacademyonline.com\/blog\/?p=591"},"modified":"2021-02-25T14:00:56","modified_gmt":"2021-02-25T08:30:56","slug":"how-to-solve-neet-2018-with-a-smile-iv","status":"publish","type":"post","link":"https:\/\/www.physicsacademyonline.com\/blog\/neet-2018\/how-to-solve-neet-2018-with-a-smile-iv.html","title":{"rendered":"How To Solve NEET 2018 With A Smile &#8211; IV"},"content":{"rendered":"<h3 style=\"font-size: 1.20em; font-weight: normal !important;\"><i>Let <em><del>Us<\/del> <\/em><\/i><span style=\"font-family: impact; letter-spacing: 1.5px; font-size: 24px; color: #000;\">YOU<\/span><i> Solve NEET 2018<br \/>\n<\/i><\/h3>\n<p>We shall discuss some more Physics questions from NEET 2018. As usual, we shall help you to link them with the relevant theories. If required, you can study the theories from the video lectures referenced @ <a href=\"https:\/\/www.physicsacademyonline.com\" target=\"_blank\" rel=\"noopener\">PhysicsAcademyOnline.com.<\/a> Back into the question, you will solve it by your own effort. Lastly, you will confirm your answer from the video solution given in the blog.<\/p>\n<p>&nbsp;<\/p>\n<p><strong><u>Question 174<\/u>:<\/strong>\u00a0 A block of mass m is placed on a smooth inclined wedge ABC of inclination <em>\u03b8<\/em> , as shown in the figure . The wedge is given an acceleration, a, towards the right. The relation between a and <em>\u03b8<\/em> \u00a0for the block to remain stationary on the wedge is:<\/p>\n<p><a href=\"https:\/\/www.physicsacademyonline.com\/blog\/wp-content\/uploads\/2019\/08\/Blog21_Img.png\"><img decoding=\"async\" loading=\"lazy\" class=\"alignnone wp-image-702\" src=\"https:\/\/www.physicsacademyonline.com\/blog\/wp-content\/uploads\/2019\/08\/Blog21_Img.png\" alt=\"\" width=\"362\" height=\"247\" srcset=\"https:\/\/www.physicsacademyonline.com\/blog\/wp-content\/uploads\/2019\/08\/Blog21_Img.png 469w, https:\/\/www.physicsacademyonline.com\/blog\/wp-content\/uploads\/2019\/08\/Blog21_Img-300x205.png 300w\" sizes=\"(max-width: 362px) 100vw, 362px\" \/><\/a><\/p>\n<p>(1) a = g \/cosec<em>\u03b8<\/em> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (2) a = g tan<em>\u03b8<\/em><em>\u00a0 <\/em>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (3) a = g cos<em>\u03b8<\/em><em>\u00a0\u00a0<\/em>\u00a0\u00a0\u00a0\u00a0\u00a0 (4) a = g \/sin<em>\u03b8<\/em><\/p>\n<p>What is the effect of the accelerated motion of the wedge on the block? Subscribe to <a href=\"https:\/\/www.physicsacademyonline.com\" target=\"_blank\" rel=\"noopener\">PhysicsAcademyOnline.com<\/a>, and learn the basics from:<span style=\"color: #8270a4;\">\u00a0 <em>Chapter Name \u2013 Mechanics; Category \u2013\u00a0 Basic; Topic Name \u2013 Newton&#8217;s Laws of Motion; Video Name \u2013 Pseudo Force.<\/em><\/span><\/p>\n<p>Show all the forces acting on the block in wedge-frame. What are their components, if any, along the incline AB? Brush up from the lecture: <em><span style=\"color: #8270a4;\"> Chapter Name \u2013 Mechanics; Category \u2013\u00a0 Basic; Topic Name \u2013 Vectors; Video Name \u2013 Resolution of Vectors.<\/span><\/em><\/p>\n<p>For what condition does the block remain stationary?<\/p>\n<p>&nbsp;<\/p>\n<p><iframe loading=\"lazy\" width=\"740\" height=\"416\" src=\"https:\/\/www.youtube.com\/embed\/k8oXSC3Lf18?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/p>\n<p>&nbsp;<\/p>\n<p><strong><u>Question 175<\/u>:<\/strong>\u00a0 A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm, and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of -0.004 cm, the correct diameter of the ball is:<\/p>\n<p>(1) 0.521 cm\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0 \u00a0 (2) 0.529 cm\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0 (3) 0.053 cm\u00a0 \u00a0 \u00a0 \u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (4) 0.525 cm<\/p>\n<p>This question is related to measurement of physical quantities in the laboratory. A simple formula can do the trick as follows:<\/p>\n<p>Actual length = main scale reading + least count x coinciding scale division \u2013 zero error<\/p>\n<p>Pick your choice, and confirm from below.<\/p>\n<p>&nbsp;<\/p>\n<p><iframe loading=\"lazy\" width=\"740\" height=\"416\" src=\"https:\/\/www.youtube.com\/embed\/Z5TgnMc4xs0?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/p>\n<p>&nbsp;<\/p>\n<p><strong><u>Question 176<\/u>:<\/strong>\u00a0 The moment of the force, <strong>F<\/strong> = (4<strong>i<\/strong> +5<strong>j<\/strong> &#8211; 6<strong>k<\/strong>) N, applied at (2, 0, -3) m, about the point (2, -2, -2) m is given by:<\/p>\n<p>(1) (- 8<strong>i<\/strong> &#8211; 4<strong>j<\/strong> &#8211; 7<strong>k<\/strong>) N-m\u00a0\u00a0\u00a0\u00a0\u00a0 (2) (- 7<strong>i<\/strong> &#8211; 4<strong>j<\/strong> &#8211; 8<strong>k<\/strong>) N-m\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0 (3) (- 7<strong>i<\/strong> &#8211; 8<strong>j<\/strong> &#8211; 4<strong>k<\/strong>) N-m\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (4) (- 4<strong>i<\/strong> &#8211; <strong>j<\/strong> &#8211; 8<strong>k<\/strong>) N-m<\/p>\n<p>What is the formula for moment (or torque) of a force about an origin? Learn from this lecture: <em><span style=\"color: #8270a4;\">Chapter Name \u2013 Mechanics; Category \u2013\u00a0 Basic; Topic Name \u2013 Rotational Mechanics; Video Name \u2013 Torque of a Force Acting on a Particle.<\/span><\/em><\/p>\n<p>What is the position vector, <strong>r<\/strong>, of the point of application of the force?\u00a0 You might need this: <span style=\"color: #8270a4;\"><em>Chapter Name \u2013 Mechanics; Category \u2013\u00a0 Basic; Topic Name \u2013 Vectors;\u00a0Video Name \u2013\u00a0 Resolution of a position vector in space.<\/em> <\/span><\/p>\n<p>How do you calculate a vector product of two vectors? See similar here: <em><span style=\"color: #8270a4;\">Chapter Name \u2013 Mechanics; Category \u2013\u00a0 Basic; Topic Name \u2013 Vectors; Video Name \u2013\u00a0 Vector product of two vectors.<br \/>\n<\/span><\/em><\/p>\n<p>&nbsp;<\/p>\n<p><iframe loading=\"lazy\" width=\"740\" height=\"416\" src=\"https:\/\/www.youtube.com\/embed\/-jlt7CK8UYY?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/p>\n<p>&nbsp;<\/p>\n<p><strong><u>Question 180<\/u>:<\/strong>\u00a0 At what temperature will the rms speed of oxygen molecules become just sufficient for escaping the Earth&#8217;s atmosphere? Given the mass of oxygen molecule, m = 5.31 x 10<sup>-26<\/sup> kg; Boltzmann&#8217;s constant, k<sub>B<\/sub> = 1.38 x 10<sup>-23<\/sup> J\/K.<\/p>\n<p>(1) 2.51 x 10<sup>4<\/sup> K\u00a0\u00a0\u00a0 \u00a0 \u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (2) 1.25 x 10<sup>4<\/sup> K\u00a0\u00a0\u00a0 \u00a0 \u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (3) 5.02 x 10<sup>4<\/sup> K\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0 \u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 (4) 1.61 x 10<sup>5<\/sup> K<\/p>\n<p>What is the formula for rms speed of the molecules of a gas? Follow the lecture: <span style=\"color: #8270a4;\">\u00a0 <em>Chapter Name \u2013 Heat and Thermodynamics; Category \u2013\u00a0 Basic; Topic Name \u2013 Kinetic Theory of Gases and Properties of Vapour; Video Name \u2013 Translational Kinetic Energy of a Gas.<\/em><\/span><\/p>\n<p>What is the escape speed from Earth&#8217;s atmosphere? You are expected to remember its value. For detailed study, however, see: <em><span style=\"color: #8270a4;\"> Chapter Name \u2013 Mechanics; Category \u2013\u00a0 Basic; Topic Name \u2013 Gravitation; Video Name \u2013 Escape Speed.<\/span><\/em><\/p>\n<p>Use the condition given, and calculate the required temperature.<\/p>\n<p>&nbsp;<\/p>\n<p><iframe loading=\"lazy\" width=\"740\" height=\"416\" src=\"https:\/\/www.youtube.com\/embed\/THjh2iU2adE?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Let Us YOU Solve NEET 2018 We shall discuss some more Physics questions from NEET 2018. As usual, we shall help you to link them with the relevant theories. If&#8230;<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[27],"tags":[],"_links":{"self":[{"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/posts\/591"}],"collection":[{"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/comments?post=591"}],"version-history":[{"count":17,"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/posts\/591\/revisions"}],"predecessor-version":[{"id":712,"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/posts\/591\/revisions\/712"}],"wp:attachment":[{"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/media?parent=591"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/categories?post=591"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/tags?post=591"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}