{"id":1584,"date":"2020-07-10T09:31:46","date_gmt":"2020-07-10T09:31:46","guid":{"rendered":"https:\/\/murreeroad.org\/physics10\/?p=1584"},"modified":"2020-10-26T00:07:11","modified_gmt":"2020-10-25T19:07:11","slug":"physics-10-chapter-14-examples","status":"publish","type":"post","link":"https:\/\/murreeroad.org\/physics10\/physics-10-chapter-14-examples\/","title":{"rendered":"Physics 10 Chapter 14 Examples"},"content":{"rendered":"

<\/a>Chapter 14 Numerical Problems Examples<\/span><\/strong><\/p>\n

Example 14.1: <\/span>If 0.5 C charge passes through a wire in 10 s, then what will be the value of current flowing through the wire?<\/strong>
\nSolution:<\/span> <\/strong>Given that, Q = 0.5 C, t= 10 s, therefore by using
\nI = Q\/t = 0.5 C\/10 s=0.05 A= 50 mA<\/p>\n

Example 14.2:<\/span> Reading on voltmeter connected across a heating element is 60 V. The amount of current passing through the heating element measured by an ammeter is 2 A.<\/strong>
\nWhat is the resistance of the heating coil of the element?<\/strong>
\nSolution:<\/span> <\/strong>Given that, V = 60 V, I = 2 A
\nUsing Ohm’s law
\nV = IR<\/p>\n

or R = V\/I = 60V\/2A 30 V A-1<\/sup> = 30 \u2126<\/p>\n

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Example 14.6:<\/span> If a current of 0.5 A passes through a bulb connected across a battery of 6 for 20 seconds, then find V<\/strong>
\nthe rate of energy transferred to the bulb. Also find the resistance of the bulb.<\/strong>
\nSolution:<\/span> <\/strong>Given that, I = 0.5 A, V=6 V, t = 20 s
\nNow using the formula,
\nEnergy W = VIt
\nwe get, Energy = 6 V \u00d7 0.5 A \u00d7 20 s = 60 J
\nSo the rate of energy transferred must be 60 J in 20 s or 3 J s-1 <\/sup>or 3 watt.<\/p>\n

Now using, Energy = W = I2<\/sup> Rt
\nWe get resistance as<\/p>\n

3 = (0.5)2<\/sup> \u00d7 R \u00d7 20
\nR = 3 \u00d71\/20 \u00d7 1\/0.25 = 3\/5 = 0.6 \u2126<\/p>\n

Example 14.7:<\/span> The resistance of an electric bulb is 500 \u03a9. Find the power consumed by the bulb when a potential difference<\/strong>
\nof 250 V is applied across its ends.<\/strong>
\nSolution:<\/span> <\/strong>Given that, R = 500 \u03a9, V = 250 V
\nUsing the formula, I = V\/R
\nI = 250 V\/ 500 = 0.5 A \u03a9
\nand Power P = I2<\/sup> R = (0.5 A)2<\/sup> \u00d7 500 \u03a9 = 125 W<\/p>\n

Example 14.8:<\/span> Calculate the one month cost of using 50 W energy saver for 8 hours daily in your study room. Assume<\/strong>
\nthat the price of a unit is Rs. 12.<\/strong>
\nSolution: <\/span><\/strong>Given that, Power = 50 W = 0.05 kW, time = 8 hours
\nNumber of units consumed = 8 \u00d7 30 \u00d7 0.05 =12 units
\nTherefore, total cost = 12 \u00d7 12 = Rs. 144<\/p>\n

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Chapter 14 Numerical Problems Examples Example 14.1: If 0.5 C charge passes through a wire in 10 s, then what will be the value of current flowing through the wire? Solution: Given that, Q = 0.5 C, t= 10 s, therefore by using I = Q\/t = 0.5 C\/10 s=0.05 A= 50 mA Example 14.2:… Read More »Physics 10 Chapter 14 Examples<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"neve_meta_sidebar":"","neve_meta_container":"","neve_meta_enable_content_width":"off","neve_meta_content_width":0,"neve_meta_title_alignment":"","neve_meta_author_avatar":"","neve_post_elements_order":"","neve_meta_disable_header":"","neve_meta_disable_footer":"","neve_meta_disable_title":""},"categories":[9],"tags":[],"_links":{"self":[{"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/posts\/1584"}],"collection":[{"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/comments?post=1584"}],"version-history":[{"count":7,"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/posts\/1584\/revisions"}],"predecessor-version":[{"id":2100,"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/posts\/1584\/revisions\/2100"}],"wp:attachment":[{"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/media?parent=1584"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/categories?post=1584"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/tags?post=1584"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}